\filename{ma310ex.tex}
\versiondate{11.3.02}
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
{\lower-.3ex\hbox{$\ssbullet$}}
{\ssbullet}
{\ssbullet}}
\def\cycle#1{(\overleftarrow{#1})}
\def\Orb{\mathop{\text{Orb}}}
\def\Stab{\mathop{\text{Stab}}}
\def\kernel{\mathop{\text{ker}}}
\def\dbhfill{\penalty-100\discretionary{}{}{}\hfill}
\def\ttablerule{\ \vrule\ }
\def\qnAA{\fixnumber\qnnumberAA\qnAAdonetrue
(a) Define {\it semigroup}. Explain what is meant by an identity in a semigroup and an inverse of an element in a semigroup with identity.
(b) Show that identities and inverses, if they exist at all, are unique.
(c) Let $F$ be a semigroup. Suppose that there is an $e\in F$ such that $xe=x$ for every $x\in F$, and for every $x\in F$ there is a
$y\in F$ such that $xy=e$. Show that
\quad(i) for every $x\in F$ there is a $z\in F$ such that $ez=x$;
\quad(ii) $e$ is an identity of $F$;
\quad(iii) $F$ is a group.
(d) Let $F$ be a non-empty semigroup. Suppose that for all $x$, $y\in F$ there are $z$, $z'\in F$ such that $xz=z'x=y$. Show that
\quad(i) if $x_0$, $e\in F$ are such that $x_0e=x_0$, then $xe=x$ for every $x\in F$;
\quad(ii) $F$ is a group
(e) Let $F$ be a non-empty finite semigroup which satisfies both cancellation laws (i.e., $xz=yz\Rightarrow x=y$ and $zx=zy\Rightarrow x=y$). Show that $F$ is a group.
\medskip
\noindent(This question is quite hard. But each fragment is supposed to follow without an impossibly large jump from what has gone before. So even if you can't do (c-i), you ought to be able to use it to manage (c-ii).)
\dbhfill[AA]}
\def\ansAA{\fixnumber\qnnumberAA\qnAAdonetrue
(a) A {\bf semigroup} is a set $F$ together with a binary operation $(x,y)\mapsto xy:F\times F\to F$ such that $(xy)z=x(yz)$ for all $x$, $y$,
$z\in F$. An {\bf identity} if $F$ is an element $e$ of $F$ such that $xe=ex=x$ for every $x\in F$. If $F$ has an identity $e$, and $x\in F$, an {\bf inverse} of $x$ is a $y\in F$ such that $xy=yx=e$.
\medskip
(b)(i) If $e$, $e'$ are identities in $F$, then $e=ee'=e'$.
\medskip
\quad(ii) If $y$, $y'$ are inverses of $x\in F$, then $y=ye=y(xy')=(yx)y'=ey'=y'$.
\medskip
(c)(i) There is a $y\in F$ such that $xy=e$, and now there is a $z\in F$ such that $yz=e$; now $ez=(xy)z=x(yz)=xe=x$.
\medskip
\quad(ii) If $x\in F$ there is a $z\in F$ such that $ez=x$; now $ex=e(ez)=(ee)z=ez=x=xe$. Thus $e$ is an identity in $F$.
\medskip
\quad(iii) If $x\in F$ we know that there are $y$, $z\in F$ such that $xy=yz=e$ and $ez=x$, as in (i) above. But (ii) now tells us that $z=ez=x$, so that $xy=yx=e$ and $y$ is an inverse of $x$ in $F$.
Thus every element of $F$ has an inverse, and $F$ is a group.
\medskip
(d)(i) If $x\in F$, there is a $z\in F$ such that $x=zx_0$, so that $xe=(zx_0)e=z(x_0e)=zx_0=x$.
\quad(ii) Because $F$ is not empty, it does have an element, call this $x_0$. We know that there is an $e\in F$ such that $x_0e=x_0$, and now (by (i)) $xe=x$ for every $x\in F$. Also, for every $x\in F$, there is a $y\in F$ such that $xy=e$. By (c), $F$ is a group.
\medskip
(e) Take $x$, $y\in F$. The map $z\mapsto xz:F\to F$ is injective. As $F$ is finite, it must also be surjective. So there is a $z\in F$ such that $xz=y$.
Similarly, because $z'\mapsto z'x:F\to F$ is injective, there is a $z'\in F$ such that $z'x=y$.
By (d), this is enough to show that $F$ is a group.
}%end of ansAA
\def\qnAB{For integers $n\ge 2$ write $\Bbb Z_n=\{0,1,\ldots,n-1\}$ and let $\times_n$ be multiplication mod $n$ on $\Bbb Z_n$, so that $3\times_75=1$.
(a) Show that $(\Bbb Z_n,\times_n)$ is a semigroup with identity, for any
$n\ge 2$.
(b) Show that an element $i$ of $\Bbb Z_n$ is invertible, for $\times_n$, iff $i$ and $n$ have no common factor other than $1$. (You may assume the fact that there are integers $r$ and $s$ such that $ir+ns$ is a factor of both $i$ and $n$.)
(c) Write $\Bbb Z^*_n$ for the group of invertible elements of $\Bbb Z_n$ (for $\times_n$). Which of the groups $\Bbb Z^*_n$, for $2\le n\le 12$, are cyclic?
Find generators of those which are.
\dbhfill[AB]}
\def\ansAB{(a) If $i$, $j$, $k\in\Bbb Z_n$ then
\Centerline{$i\times_n(j\times_nk)\equiv_ni(j\times_nk)\equiv_ni(jk)
=(ij)k\equiv_n(i\times_nj)k\equiv_n(i\times_nj)\times_nk$;}
\noindent but both $i\times_n(j\times_nk)$ and $(i\times_nj)\times_nk$ belong to $\{0,\ldots,n-1\}$, so if they differ by a multiple of $n$ they must be equal.
Thus $(\Bbb Z_n,\times_n)$ is a semigroup.
Since $i\times_n1=1\times_ni=i$ for every $i\in\Bbb Z_n$, $1$ is an identity of $\Bbb Z_n$.
\medskip
(b)(i) If $i\in\Bbb Z_n$ is invertible, with inverse $j$ say, then $ij-1$ is a multiple of $n$; now any common factor of $i$ and $n$ is also a factor of $ij$ and $ij-1$ so must be $1$.
\medskip
\quad(ii) If $i\in\Bbb Z_n$ and $n$ have no non-trivial common factor, let $r$, $s\in\Bbb Z$ be such that $ri+sn$ is a factor of both $i$ and $n$. Then $ri+sn$ must actually be $1$, so $ri\equiv_n1$. Let $j\in\Bbb Z_n$ be such that $r\equiv_nj$; then $ij\equiv_n1$ so $j$ is an inverse of $i$ in $\Bbb Z_n$ and $i$ is invertible.
\medskip
(d) Tabulating,
$$\vbox{\offinterlineskip
\halign{\qquad\hfil##\quad\ttablerule\quad&\hfil##\hfil\quad\ttablerule
&\hfil##\hfil
\quad&##\hfil\strut\cr
$n$&$\Bbb Z_n^*$&generators&\cr
\noalign{\hrule}
2&$\{1\}$&$1$\cr
3&$\{1,2\}$&$2$\cr
4&$\{1,3\}$&$3$\cr
5&$\{1,2,3,4\}$&2, 3\cr
6&$\{1,5\}$&5\cr
7&$\{1,2,3,4,5,6\}$&3, 5\cr
8&$\{1,3,5,7\}$&--&$\Bbb Z_8^*\cong K_4$\cr
9&$\{1,2,4,5,7,8\}$&2, 5\cr
10&$\{1,3,7,9\}$&3, 7\cr
11&$\{1,2,3,4,5,6,7,8,9,10\}$&2, 6, 7, 8\cr
12&$\{1,5,7,11\}$&--&$\Bbb Z_{12}^*\cong K_4$\cr
}}$$
}%end of ansAB
\def\qnBA{(a) Define {\it subgroup}, {\it coset}.
(b) State and prove Lagrange's theorem.
(c) What is meant by `the subgroup generated by' an element of a group? Describe the subgroup of the group of non-singular $2\times 2$ real matrices generated by $\Matrix{1&2\\0&1}$.
(d) Show that if $G$ is a finite group of order $n$ and $a\in G$ then $a^n=e$.
\dbhfill[BA]}
\def\ansBA{(a) Let $G$ be a group.
\medskip
\quad(i) A {\bf subgroup} of $G$ is a set $H\subseteq G$ such that ($\alpha$) $xy\in H$ for all $x$, $y\in H$ ($\beta$) $e\in H$ ($\gamma$) $x^{-1}\in H$ for every $x\in H$.
\medskip
\quad(ii) For $A\subseteq G$, a {\bf left coset} of $A$ is a set of the form $xA=\{xa:a\in A\}$ where $x\in G$; a {\bf right coset} of $A$ is a set of the form $Ax=\{ax:a\in A\}$ where $x\in G$.
\medskip
(b) {\bf Lagrange's theorem} If $G$ is a finite group and $H$ is a subgroup of $G$ then $\#(H)$ is a factor of $\#(G)$.
\Prf\ (i) If $x\in G$ then the function $x\mapsto xh:H\to xH$ is injective (by the cancellation law) and surjective (by the definition of $xH$), so $\#(xH)=\#(H)$.
(ii) If $y\in G$, then $yH$ is a left coset of $H$ containing $ye=y$.
On the other hand, if we have any left coset $V$ of $H$ containing $y$, then $V=xH$ for some $x\in H$. Now $yH\subseteq xHH\subseteq xH=V$; but also there is some $h\in H$ such that $y=xh$, so that $x=yh^{-1}\in yH$ and
$V=xH\subseteq yHH\subseteq yH$. Accordingly $V=yH$.
What this means is that every member of $G$ belongs to exactly one left coset of $H$.
(iii) Since all the left cosets of $H$ have the same number $\#(H)$ of elements,
\Centerline{$\#(G)=$(number of left cosets of $H$)$\times \#(J)$}
\noindent is a multiple of $\#(H)$.\ \Qed
\medskip
(c) If $G$ is a group and $a\in G$, the {\bf subgroup generated by} $a$ is $=\{a^n:n\in\Bbb Z\}$, the smallest subgroup of $G$ containing $a$.
The subgroup generated by $\Matrix{1&2\\0&1}$ is
$\{\Matrix{1&2n\\0&1}:n\in\Bbb Z\}$.
\medskip
(d) Examine $e,a,a^2,a^3,\ldots$. Because $G$ is finite, these cannot all be different. Let $r\in\Bbb N\setminus\{0\}$ be the least number such that $a^r=a^k$ for some $0\le k$.
The subgroup is therefore a non-abelian group with six elements and must be isomorphic to $S_3$.
\medskip
(b)(i) If $H$ is a subgroup of $\Bbb C\setminus\{0\}$ with $n$ elements, then every element of $H$ is an $n$th root of $1$. But the $n$th complex roots of $1$ in $\Bbb C$ are just the powers of $e^{2\pi i/n}$, so form a cyclic group.
Conversely, any finite cyclic group with $n$ elements is isomorphic to the subgroup $$ of $\Bbb C\setminus\{0\}$.
\medskip
(ii) The only finite subgroups of $\Bbb R\setminus\{0\}$ are $\{1\}\cong\Bbb Z_1$ and $\{1,-1\}\cong\Bbb Z_2$.
}%end of ansBB
\def\qnBC{(a) Define {\it normal subgroup}.
(b) Explain how to form the quotient group of a given group by a normal subgroup. At which point does the construction fail if the subgroup is not normal?
(c) Let $G$ be the group of non-singular $2\times 2$ real matrices. Describe (with proofs)
\quad(i) a normal subgroup $H$ of $G$ which is neither $\{e\}$ nor $G$;
\quad(ii) a familiar group isomorphic to $G/H$;
\quad(iii) a subgroup $K$ of $G$ which is not normal.
\dbhfill[BC]}
\def\ansBC{(a) A {\bf normal subgroup} of $G$ is a subgroup $H$ of $G$ such that $gH=Hg$ for every $g\in G$.
\medskip
(b) If $H$ is a normal subgroup of $G$, let $G/H$ be the set of (left or right) cosets of $H$ in $G$. Then $AB\in G/H$ for all $A$, $B\in G/H$. \Prf\ We can express $A$ as $aH$ and $B$ as $bH$. Now
\Centerline{$AB=(aH)(bH)=a(Hb)H=a(bH)H=abHH=abH\in G/H$. \Qed}
\noindent So we have a binary operation $(A,B)\mapsto AB:G/H\times G/H\to G/H$.
If $A$, $B$, $C\subseteq G$, then $(AB)C=A(BC)$; in particular, $(AB)C=A(BC)$ for all $A$, $B$, $C\in G/H$, so $G/H$ is a semigroup. $H=eH$ belongs to $G/H$, and $eH\cdot aH=(ea)H=aH=(ae)H=aH\cdot eH$ for every $a\in G$, so $eH$ is an identity in the semigroup $G/H$. If $a\in G$ then $a^{-1}H\in G/H$ and
\Centerline{$a^{-1}H\cdot aH=aa^{-1}H=eH=a^{-1}aH=a^{-1}H\cdot aH$,}
\noindent so $a^{-1}H$ is an inverse of $aH$ in $G/H$; thus every element of $G/H$ has an inverse and $G/H$ is a group.
If $H$ is not normal then $aHbH$ need not belong to $G/H$ for all $a$, $b\in G$ so we can't use this as a definition of multiplication in $G/H$.
\medskip
(c)(i) Try $H=\{\tbf{A}:\tbf{A}\in G,\,\det\tbf{A}=1\}$. Because $\det:G\to(\Bbb R\setminus\{0\},\times)$ is a homomorphism, its kernel $H$ is a normal subgroup of $G$.
\medskip
\quad(ii) $G/H$ is isomorphic to the image
$\{\det\tbf{A}:\tbf{A}\in G\}=\Bbb R\setminus\{0\}$.
\medskip
\quad(iii) Try $K=\{\Matrix{1&\alpha\\0&1}:\alpha\in\Bbb R\}$. Then $K$ is a subgroup of $G$ because it is the image of the homomorphism $\alpha\mapsto\Matrix{1&\alpha\\0&1}:(\Bbb R,+)\to G$. It is not normal because, e.g.,
\Centerline{$\Matrix{0&1\\1&0}\Matrix{1&1\\0&1}\Matrix{0&1\\1&0}
=\Matrix{1&0\\1&0}\notin K$}
\noindent even though $\Matrix{0&1\\1&0}^{-1}=\Matrix{0&1\\1&0}$ and $\Matrix{1&1\\0&1}\in K$.
}%end of ansBC
\def\qnBD{\fixnumber\qnnumberBD\qnBDdonetrue
$A_4$ is the group of permutations $e$=identity,
$a=\cycle{1 2}\cycle{3 4}$,
$b=\cycle{1 3}\cycle{2 4}$ $c=\cycle{1 4}\cycle{2 3}$, $p=\cycle{1 2 3}$,
$q=\cycle{2 4 3}$, $r=\cycle{1 4 2}$, $s=\cycle{1 3 4}$, $w=\cycle{1 3 2}$,
$x=\cycle{1 4 3}$, $y=\cycle{2 3 4}$, $z=\cycle{1 2 4}$.
(a) Write out the group table of $A_4$, listing the elements in the order given both in rows and in columns.
(b) Find a normal subgroup $H$ with four elements; list its cosets, and give the group table of $A_4/H$.
(c) How many subgroups with three elements are there? Are any of them normal?
(d) Is there any subgroup with six elements?
\dbhfill[BD]}
\def\ansBD{\fixnumber\qnnumberBD\qnBDdonetrue
(a)
$$\vbox{\offinterlineskip
\halign{\hfil##\hfil&\quad\ttablerule\hfil##\hfil&\quad\hfil##\hfil
&\quad\hfil##\hfil
&\quad\hfil##\hfil&\quad\hfil##\hfil&\quad\hfil##\hfil&\quad\hfil##\hfil
&\quad\hfil##\hfil&\quad\hfil##\hfil
&\quad\hfil##\hfil&\quad\hfil##\hfil&\quad\hfil##\hfil\strut\cr
&$e$&$a$&$b$&$c$&$p$&$q$&$r$&$s$&$w$&$x$&$y$&$z$\cr
\noalign{\hrule}
$e$&$e$&$a$&$b$&$c$&$p$&$q$&$r$&$s$&$w$&$x$&$y$&$z$\cr
$a$&$a$&$e$&$c$&$b$&$q$&$p$&$s$&$r$&$x$&$w$&$z$&$y$\cr
$b$&$b$&$c$&$e$&$a$&$r$&$s$&$p$&$q$&$y$&$z$&$w$&$x$\cr
$c$&$c$&$b$&$a$&$e$&$s$&$r$&$q$&$p$&$z$&$y$&$x$&$w$\cr
$p$&$p$&$s$&$q$&$r$&$w$&$z$&$x$&$y$&$e$&$c$&$a$&$b$\cr
$q$&$q$&$r$&$p$&$s$&$x$&$y$&$w$&$z$&$a$&$b$&$e$&$c$\cr
$r$&$r$&$q$&$s$&$p$&$y$&$x$&$z$&$w$&$b$&$a$&$c$&$e$\cr
$s$&$s$&$p$&$r$&$q$&$z$&$w$&$y$&$x$&$c$&$e$&$b$&$a$\cr
$w$&$w$&$y$&$z$&$x$&$e$&$b$&$c$&$a$&$p$&$r$&$s$&$q$\cr
$x$&$x$&$z$&$y$&$w$&$a$&$c$&$b$&$e$&$q$&$s$&$r$&$p$\cr
$y$&$y$&$w$&$x$&$z$&$b$&$e$&$e$&$c$&$r$&$p$&$q$&$s$\cr
$z$&$z$&$x$&$w$&$y$&$c$&$a$&$a$&$b$&$s$&$q$&$p$&$r$\cr
}}$$
\medskip
(b) $H=\{e,a,b,c\}$ is the normal subgroup with 4 elements; its cosets are
$A=\{p,q,r,s\}$ and $B=\{w,x,y,z\}$; the group table is
$$\vbox{\halign{\hfil##\hfil&\ttablerule\hfil##\hfil&\hfil##\hfil&\hfil##\hfil
\strut\cr
&$H$&$A$&$B$\cr
\noalign{\hrule}
$H$&$H$&$A$&$B$\cr
$A$&$A$&$B$&$H$\cr
$B$&$B$&$H$&$A$\cr}}$$
\medskip
(c) There are four subgroups with three elements ($\{e,p,w\}$, $\{e,q,y\}$, $\{e,r,z\}$, $\{e,s,x\}$); none are normal.
\medskip
(d) There is no subgroup with six elements.
}%end of ansBD
\def\qnBE{(a) Define {\it homomorphism}, {\it isomorphism}, {\it kernel} (of a homomorphism).
(b) Show that the kernel of a homomorphism is a normal subgroup and that the set of values of a homomorphism is a subgroup.
(c) State and prove the First Isomorphism Theorem.
(d) Let $G$ be a group, $H$ a normal subgroup of $G$ and $K$ a subgroup of $G$. Show that
\quad(i) $HK$ is a subgroup of $G$;
\quad(ii) $H\cap K$ is a normal subgroup of $K$;
\quad(iii) $HK/H$ is isomorphic to $K/H\cap K$.
\dbhfill[BE]}
\def\ansBEa{(a) If $G$ and $H$ are groups, a homomorphism from $G$ to $H$ is a function $\phi:G\to H$ such that $\phi(ab)=\phi(a)\phi(b)$ for all $a$,
$b\in G$. $\phi$ is an isomorphism if it is a bijective homomorphism. The kernel of $\phi$ is $\{a:a\in G,\,\phi(a)=e_H\}$.
\medskip
(b)(i) Let $\phi:G\to H$ be a homomorphism and $K$ its kernel. Then $e_G\in K$ because $\phi(e_G)=e_H$. If $a$, $b\in K$, then
\Centerline{$\phi(ab)=\phi(a)\phi(b)=e_He_H=e_H$,}
\noindent so $ab\in K$. If $a\in K$, then
\Centerline{$\phi(a^{-1})=\phi(a)^{-1}=e_H^{-1}=e_H$,}
\noindent so $a^{-1}\in K$. Thus $K$ is a subgroup of $G$. If $a\in G$ and $k\in K$ then
\Centerline{$\phi(aka^{-1})=\phi(a)\phi(k)\phi(a^{-1})
=\phi(a)\cdot e_H\cdot\phi(a)^{-1}
=\phi(a)\phi(a)^{-1}=e_H$,}
\noindent so $aka^{-1}\in K$; as $a$ and $k$ are arbitrary,
$K\normalsubgroup G$.
\medskip
\quad(ii) Let $\phi:G\to H$ be a homomorphism and $K$ its set of values. Then $e_H\in K$ because $e_H=\phi(e_G)$. If $x$, $y\in K$ there are $a$, $b\in G$ such that $\phi(a)=x$ and $\phi(b)=y$; now
\Centerline{$xy=\phi(a)\phi(b)=\phi(ab)\in K$.}
\noindent If $x\in K$ there is an $a\in G$ such that $\phi(a)=x$, and now
\Centerline{$x^{-1}=\phi(a)^{-1}=\phi(a^{-1})\in K$.}
\noindent Thus $K$ is a subgroup of $G$.
}%end of ansBEa
\def\ansBEc{(c) {\bf First Isomorphism Theorem} Let $G$ and $H$ be groups, and $\phi:G\to H$ a homomorphism. Write $K$ for the kernel of $\phi$ and $\phi[G]$ for the set of values of $\phi$, so that $K$ is a normal subgroup of $G$ and $\phi[G]$ is a subgroup of $H$. Then the quotient group $G/K$ is isomorphic to $\phi[G]$.
\medskip
\noindent{\bf proof} (i) For $x$, $y\in G$, $\phi(x)=\phi(y)$ iff $xK=yK$.
\Prf\ ($\alpha$) If $xK=yK$, then $y\in xK$ so there is a $k\in K$ such that $y=xk$ and
\Centerline{$\phi(y)=\phi(xk)=\phi(x)\phi(k)=\phi(x)e_H=\phi(x) $.}
\noindent ($\beta$) If $\phi(x)=\phi(y)$, set $k=x^{-1}y$; then
\Centerline{$\phi(k)=\phi(x)^{-1}\phi(y)=e_H$,}
\noindent and $k\in K$. But $y=xk$, so $y\in xK$; as certainly $y\in yK$,
$xK$ and $yK$ have a common element and must be identical.\ \Qed
\medskip
\quad(ii) We can therefore define $\psi:G/K\to\phi[G]$ by saying that $\psi(A)=\phi(x)$ whenever $x\in G$ and $A=xK$. If $A$, $B\in G/H$, express them as $A=xK$ and $B=yK$ where $x$, $y\in G$; then $AB=xyK$, so
\Centerline{$\psi(AB)=\phi(xy)=\phi(x)\phi(y)=\psi(A)\psi(B)$.}
\noindent Thus $\psi:G/H\to\phi[G]$ is a homomorphism. It is injective because if $A$, $B\in G/H$ and $\psi(A)=\psi(B)$, we can express $A$ as $xK$ and $B$ ans $yK$, and $\phi(x)=\psi(A)=\psi(B)=\phi(y)$, so $A=xK=yK=B$. It is surjective because if $h\in\phi[H]$ there is some $x\in G$ such that $h=\phi(x)=\psi(xK)$.
Thus $\psi$ is a bijective homomorphism and is an isomorphism.
\medskip
(d)(i) $e=ee\in HK$. If $x$, $y\in HK$ we can express them as $x=h_1k_1$, $y=h_2k_2$ where $h_1$, $h_2\in H$ and $k_1$, $k_2\in K$; now
\Centerline{$xy=h_1k_1h_2k_2\in h_1k_1Hk_2=h_1Hk_1k_2\in HHKK=HK$.}
\noindent If $x\in HK$ we can express it as $x=hk$, where $h\in H$ and $k\in K$; now
\Centerline{$x^{-1}=k^{-1}h^{-1}\in k^{-1}H=Hk^{-1}\subseteq HK$.}
\noindent Thus $HK$ is a subgroup of $G$.
\medskip
\quad(ii) $e\in H$ and $e\in K$ so $e\in H\cap K$. If $x$, $y\in H\cap K$ then $xy\in H$ and $xy\in K$ so $xy\in H\cap K$. If $x\in H\cap K$ then
$x^{-1}\in H$ and $x^{-1}\in K$ so $x^{-1}\in H\cap K$. Thus $H\cap K$ is a subgroup of $K$.
If $x\in K$ and $y\in H\cap K$ then $xyx^{-1}\in H$ (because
$H\normalsubgroup G$ and $x\in G$ and $y\in H$), while $xyx^{-1}\in K$ because $K$ is a subgroup and both $x$ and $y$ belong to $K$. So
$xyx^{-1}\in H\cap K$. As $x$ and $y$ are arbitrary,
$H\cap K\normalsubgroup K$.
\medskip
\quad(iii) (Because $H$ is a normal subgroup of $G$ it is also a normal subgroup of $HK$.) Define $\phi:K\to HK/H$ by setting $\phi(a)=aH$ for $a\in K$.
Since $aH\cdot bH=abH$ for all $a$, $b\in HK$, $\phi$ is a homomorphism. For $k\in K$,
\Centerline{$\phi(k)=e_{HK/H}\iff kH=H\iff k\in H\iff k\in H\cap K$,}
\noindent so the kernel of $\phi$ is $H\cap K$. (This gives an alternative proof that $H\cap K\normalsubgroup K$.) Also
$\phi[K]=HK/H$. \Prf\ If $A\in HK/H$, it is of the form $aH$ for some
$a\in HK$. Now $a$ is expressible as $hk$ where $h\in H$ and $k\in K$, so that
\Centerline{$aH=Ha=Hhk=Hk=kH=\phi(k)\in\phi[K]$.}
\noindent As $A$ is arbitrary, $\phi$ is surjective.\ \Qed
By the First Isomorphism Theorem
\Centerline{$K/H=\phi[K]\cong K/\kernel\phi=K/H\cap K$,}
\noindent as required.
}%end of ansBE
\def\qnBF{ Let $G$ be a group and $H$, $K$ subgroups of $G$.
(a) Show that $H\cap K$ is a subgroup of $H$.
(b) Show that if $HK\subseteq KH$ then $KH$ is a subgroup of $G$ and $HK=KH$.
(c) Show that if $H$ and $K$ are finite then $\#(HK)\#(H\cap K)=\#(H)\#(K)$.
(Hint: show that for $h_1$, $h_2\in H$, $h_1K=h_2K$ iff
$h_1(H\cap K)=h_2(H\cap K)$ iff $h_1^{-1}h_2\in K$.)
(d) Give an example in which $HK$ is not a subgroup of $G$.
(Hint: take $G=A_4$, $\#(H)=3$, $\#(K)=2$.)
\dbhfill[BF]}
\def\ansBF{(a) $e\in H$ and $e\in K$ so $e\in H\cap K$. If $x$, $y\in H\cap K$ then $xy\in H$ and $xy\in K$ so $xy\in H\cap K$. If $x\in H\cap K$ then
$x^{-1}\in H$ and $x^{-1}\in K$ so $x^{-1}\in H\cap K$. Thus $H\cap K$ is a subgroup of $G$.
\medskip
(b) $e=ee\in KH$. If $a$, $b\in KH$ then
\Centerline{$ab\in (KH)(KH)=K(HK)H=KKHH=KH$.}
\noindent If $a\in KH$, express it as $kh$ where $k\in K$, $h\in H$; then
\Centerline{$a^{-1}=h^{-1}k^{-1}\in HK\subseteq KH$.}
\noindent So $KH$ is a subgroup of $G$. Now if $a\in KH$, $a^{-1}\in KH$; express $a^{-1}$ as $kh$ where $k\in K$ and $h\in H$; then
$a=h^{-1}k^{-1}\in HK$. This shows that $KH\subseteq HK$ so $KH=HK$.
\medskip
(c) If $h_1$, $h_2\in H$, then
\Centerline{$h_1K=h_2K\iff h_2\in h_1K\iff h_1^{-1}h_2\in h_1^{-1}h_1K=K$,}
\noindent while similarly $h_1(H\cap K)=h_2(H\cap K)$ iff
$h_1^{-1}h_2\in H\cap K$. But of course $h_1^{-1}h_2\in H$, so
$h_1^{-1}h_2\in H\cap K$ iff $h_1^{-1}h_2\in K$, and $h_1(H\cap K)=h_2(H\cap K)$ iff $h_1K=h_2K$.
What this means is that we have a bijection between the set $P=\{hK:h\in H\}$ and the set $Q=\{h(H\cap K):h\in H\}$, matching $hK$ with $h(H\cap K)$ for every
$h\in H$.
Now consider $HK$. This is just the union of the cosets $hK$ with $h\in H$.
Since these all have exactly $\#(K)$ members, $\#(HK)=\#(P)\#(K)$. On the other hand, we know from Lagrange's theorem that $\#(H)=\#(Q)\#(H\cap K)$.
Since $\#(P)=\#(Q)$,
\Centerline{$\#(HK)\#(H\cap K)
=\#(P)\#(K)\#(H\cap K)=\#(Q)\#(K)\#(H\cap K)=\#(H)\#(K)$.}
\medskip
(d) We know that $A_4$ has subgroups $H=\{e,\cycle{1 2 3},\cycle{1 3 2}\}$ and
$K=\{e,\cycle{1 2}\}$. By (c), $HK$ has 6 members; but we also know that $A_4$ has no subgroup of size 6.
}%end of ansBF
\def\qnBG{(a) For $k=1,2,\ldots$, how many elements of $A_4$ have $k$ orbits?
(Hint: question \number\qnnumberBD.)
(b) For $k=1,2,\ldots$ how many elements of $S_4$ have $k$ orbits?
(You should be able to work out how many of each kind of permutation there are without listing them, but if in doubt do the exhaustive check.)
(c) Suppose that $\rho\in S_n$, where $n\ge 3$, has $m$ orbits. (i) How many orbits does $\cycle{1 2 3}\rho$ have if $1$, $2$, $3$ all belong to different orbits of $\rho$? (ii) How many orbits does $\cycle{1 2 3}\rho$ have if $1$, $2$ belong to the same orbit of $\rho$ and $3$ belongs to a different one?
(iii) Can we determine how many orbits $\cycle{1 2 3}\rho$ has if we are told
that $1$, $2$, $3$ all belong to the same orbit of $\rho$? Justify your answers.
\dbhfill[BG]}
\def\ansBG{(a) $1$ element with 4 orbits, 11 elements with 2 orbits.
\medskip
(b) 1 identity with 4 orbits, 6 transpositions with 3 orbits (because $6=\,^2C_4$), 11 elements with 2 orbits (3 double transpositions, 8 3-cycles),
6 4-cycles with 1 orbit (because there are $3!$ ways of writing a 4-cycle starting with 1).
\medskip
(c)(i) If $\rho=\cycle{1 . . .}\cycle{2 * * *}\cycle{3 ! ! !}\ldots$, then
$\cycle{1 2 3}\rho=\cycle{1 . . . 2 * * * 3 ! ' !}\ldots$ has $m-2$ orbits.
(ii) If $\rho=\cycle{1 . . . 2 * * *}\cycle{3 ! ! !}\ldots$, then
$\cycle{1 2 3}\rho=\cycle{1 . . . 3 ! ! !}\cycle{2 * * *}\ldots$ has $m$ orbits.
(iii) If $\rho=\cycle{1 . . . 2 * * * 3 ! ! !}\ldots$, then
$\cycle{1 2 3}\rho=\cycle{1 . . . 3 ! ! ! 2 * * *}\ldots$ has $m$ orbits;
but if $\rho=\cycle{1 . . . 3 ! ! ! 2 * * *}\ldots$, then
$\cycle{1 2 3}\rho=\cycle{1 . . .}\cycle{2 * * *}\cycle{3 ! ! !}\ldots$ has
$m+2$ orbits. So no, we cannot determine how many orbits $\cycle{1 2 3}\rho$ has. (Or just say: take $n=3$, $m=1$, $\rho=\cycle{1 2 3}$ or
$\cycle{1 3 2}$.)
}%end of ansBG
\def\qnBH{(a) In the group $S_6$, set $\rho=\cycle{1 2}\cycle{3 4 5 6}$. What is the order of $\rho$? List the members of the subgroup generated by $\rho$.
(b) In the group $S_7$, set $\rho=\cycle{1 2 3}\cycle{4 5 6 7}$. What is the order of $\rho$?
(c) In the group $S_8$, what is the largest order which any element can have? Why?
\dbhfill[BH]}
\def\ansBH{(a) $<\rho>=\{e,\cycle{1 2}\cycle{3 4 5 6},\cycle{3 5}\cycle{4 6},
\cycle{1 2}\cycle{3 6 5 4}\}$; the order of $\rho$ is 4.
(b) $3\times 4 = 12$.
(c) $3\times 5=15$ for $\cycle{1 2 3}\cycle{4 5 6 7 8}$ and its conjugates. This is the best we can do because anything with a 6-, 7- or 8-cycle will have order 6, 7 or 8 respectively; something with two 4-cycles will have order 4;
something with a 4-cycle and a 3-cycle will have order 12; and if we have only
2- and 3-cycles we can't get an order above 6.
}%end of ansBH
\def\qnCA{The `16-puzzle' is a square tray with 15 numbered square pieces,
one square being left empty, thus:
$$\vbox{\offinterlineskip
\halign{\ttablerule\vrule height 12pt depth 5pt width 0pt
\hfil##\hfil\ttablerule
&\hfil##\hfil\ttablerule
&\hfil##\hfil\ttablerule
&\hfil##\hfil\ttablerule\cr
\noalign{\hrule}
1&2&3&4\cr
\noalign{\hrule}
5&6&7&8\cr
\noalign{\hrule}
9&10&11&12\cr
\noalign{\hrule}
13&14&15&\vrule height 11pt depth 4pt width 10pt\cr
\noalign{\hrule}
}}$$
\noindent Each piece is free to move up, down or sideways; so that the pieces
can be rearranged by shifting them successively into the empty space, which wanders accordingly around the tray. (For instance, the first three moves
could be 15 right, 11 down, 7 down.) Show that at most $\Bover12\cdot 16!$ positions can be achieved from the starting position. ({\it Hint\/}: if you think of the empty space as `16', find a relation between the signature of a permutation
in $S_{16}$ and the position of `16' in the tray.)
\dbhfill[CA]}
\def\ansCA{Imagine that the tray, underneath the moving pieces, is coloured black and white in a chessboard pattern; for definiteness, let us take the diagonal squares to be black, so that in the original position the `16' tile is on a black square. Each move is a transposition of two tiles, one of them the `16' tile; so after an even number of moves we shall have an even permutation and the `16' tile will be on a black square, while after an odd number of moves we shall have an odd permutation and the `16' tile will be on a white square. This means that only half the $16!$ permutations will be achievable.
(It is in fact the case that exactly half the $16!$ permutations are achievable.
But the proof of this is more difficult.)
}%end of ansCA
\def\qnCB{Two elements $x$, $y$ of a group $G$ are {\bf conjugate} in $G$ if there is an $a\in G$ such that $x=aya^{-1}$.
Let $\sigma$ be the permutation $\cycle{1 2 3}$.
(a) How many permutations are conjugate to $\sigma$ in the symmetric group $S_4$?
(b) Which permutations are conjugate to $\sigma$ in the alternating group $A_4$? (You may find your group table for $A_4$ useful.)
(c) How many permutations are conjugate to $\sigma$ in the symmetric group $S_5$?
(d) How many permutations are conjugate to $\sigma$ in the alternating group $A_5$?
\dbhfill[CB]}
\def\ansCB{(a) Since $\rho\sigma\rho^{-1}=\cycle{\rho(1)\,\rho(2)\,\rho(3)}$, the conjugates of $\sigma$ in $S_4$ are just the 3-cycles, and there are 8 of them ($4\times 3\times 2$, divide by 3 because you can start anywhere).
\medskip
(b) $\cycle{1 2 3}$ (of course), $\cycle{1 3 4}$, $\cycle{1 4 2}$,
$\cycle{2 3 4}$.
(Working this out without the help of general theory is a bit of a slog. There is a general theorem that, for any finite group $G$ and any $x\in G$,
\Centerline{$\#(G)=$(no.\ of conjugates of $x)\times$(no.\ of elements commuting with $x$).}
\noindent Now there are at least 3 elements of $A_4$ commuting with
$\cycle{1 2 3}$, viz., $e$, $\cycle{1 2 3}$ and $\cycle{1 2 3}^2=\cycle{1 3 2}$, so there can't be more than 4 conjugates.)
\medskip
(c) All the 3-cycles, and there are $5\times 4\times 3\div 3=20$ of them.
\medskip
(d) This time all 20 of them are conjugate to $\sigma$ in $A_5$. To see this, take any 3-cycle $\sigma'$. Then there is a $\rho\in S_5$ such that $\rho\sigma\rho^{-1}=\sigma'$. If $\rho\in A_5$, then certainly $\sigma'$ is conjugate to $\sigma$ in $A_5$. Otherwise, because $\sigma'$ is a 3-cycle, and we have 5 elements to play with, there is a transposition $\tau$ which is disjoint from $\sigma'$, so that $\tau\sigma'=\sigma'\tau$ and $\tau\sigma'\tau^{-1}=\sigma'$. But this means that
\Centerline{$\sigma'=\tau\rho\sigma\rho^{-1}\tau^{-1}
=(\tau\rho)\sigma(\tau\rho)^{-1}$}
\noindent is conjugate to $\sigma$ in $A_5$, because $\tau\rho$ is a product of odd permutations and belongs to $A_5$.
}%end of ansCB
\def\qnCC{(a) Let $G$ be a group, $K$ a normal subgroup of $G$ and $H$ a subgroup of $G$ such that $H\cap K=\{e\}$. Show that $x\mapsto xK:H\to G/K$ is an injective homomorphism.
(b) Now let $K$ be the normal subgroup of $S_4$ with 4 elements. How many elements does the quotient group $S_4/K$ have? Show that there is a subgroup $H$ of $S_4$, isomorphic to $S_3$, such that
$H\cap K=\{e\}$. Show that $H\cong S_4/K$.
\dbhfill[CC]}
\def\ansCC{(a) By the definition of $G/K$, $xK\cdot yK=(xy)K$ for all $x$,
$y\in G$, and therefore for all $x$, $y\in H$; so the map
$x\mapsto xK:H\to G/K$ is a homomorphism. Its kernel is
$\{x:x\in H,\,xK=K\}=H\cap K=\{e\}$, so it is injective.
\medskip
(b) $\#(S_4/K)=\Bover{24}4=6$.
Let $H$ be $\{\rho:\rho\in S_4,\,\rho(4)=4\}$; then $H$ is a subgroup of $S_4$ and can be identified with $S_3$ (if $\rho\in H$, then its restriction to $\{1,2,3\}$ belongs to $S_3$).
Now $H\cap K=\{e\}$, so the map $x\mapsto xK:H\to S_4/K$ is an injective homomorphism. Because $\#(H)=6=\#(S_4/K)$, it must also be surjective, and is an isomorphism.
}%end of ansCC
\def\qnCD{Let $G$ be a group. Set
\Centerline{$A=\{xyx^{-1}y^{-1}:x,\,y\in G\}$,}
\noindent and
\Centerline{$H=\{a_1a_2\ldots a_n:a_1,\ldots,a_n\in A\}$.}
(a) Show that (i) $a^{-1}\in A$ for every $a\in A$ (ii) $zaz^{-1}\in A$ for every $a\in A$, $z\in G$ (iii) $H$ is a normal subgroup of $G$.
(b) Show that the quotient group $G/H$ is abelian. ({\it Hint\/}: $xH=yH$ iff $y^{-1}x\in H$.)
(c) Show that if $K$ is any abelian group and $\phi:G\to K$ is a homomorphism, then the kernel of $K$ includes $H$.
(d) Find $H$ when $G$ is
\quad(i) $\Bbb Z_6$\quad (ii) $S_3$\quad (iii) $A_4$.
(e) Find $A$ when $G$ is
\quad(i) $S_3$\quad (ii) $A_4$.
\noindent($H$ is the {\bf commutator subgroup} of $G$.)
\dbhfill[CD]}
\def\ansCD{(a)(i) If $a=xyx^{-1}y^{-1}\in A$ then $a^{-1}=yxy^{-1}x^{-1}\in A$.
\medskip
\quad(ii) If $a=xyx^{-1}y^{-1}\in A$ and $z\in G$, then
$$\eqalign{zaz^{-1}
&=zxyx^{-1}y^{-1}z^{-1}=zxz^{-1}zyz^{-1}zx^{-1}z^{-1}zy^{-1}z^{-1}\cr
&=(zxz^{-1})(zyz^{-1})(zxz^{-1})^{-1}(zyz^{-1})^{-1}
\in A.\cr}$$
\medskip
\quad(iii) $e=eee^{-1}e^{-1}\in A\subseteq H$.
If $x$, $y\in H$ there are strings in $A$ with products $x$ and $y$; putting these together we get a string in $A$ with product $xy$, so $xy\in H$.
If $a_1\ldots a_n\in H$ then (using (i)) $a_n^{-1}\ldots a_1^{-1}\in H$; thus $H$ is closed under inverses.
If $a_1\ldots a_n\in H$ and $x\in G$ then (using (ii))
\Centerline{$xa_1\ldots a_nx^{-1}=xa_1x^{-1}xa_2x^{-1}\ldots xa_nx^{-1}\in H$.}
\medskip
(b) If $x$, $y\in G$ then
\Centerline{$(yx)^{-1}(xy)=x^{-1}y^{-1}xy
=x^{-1}y^{-1}(x^{-1})^{-1}(y^{-1})^{-1}\in A\subseteq H$}
\noindent so
\Centerline{$xH\cdot yH=xyH=yxH=yH\cdot xH$.}
\medskip
(c) If $\phi:G\to K$ is a homomorphism and $x$, $y\in G$, then
\Centerline{$\phi(xyx^{-1}y^{-1})=\phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}
=\phi(x) \phi(x)^{-1}\phi(y)\phi(y)^{-1}=e_K$,}
\noindent so $xyx^{-1}y^{-1}$ belongs to the kernel of $\phi$; as $x$ and $y$ are arbitrary, $A\subseteq\kernel(\phi)$; as $\kernel(\phi)$ is a subgroup, $H\subseteq\kernel(\phi)$.
\medskip
(d)(i) $A=H=\{0\}$, because $Z_6$ is abelian.
\medskip
\quad(ii) $H$ must be either $\{e\}$ or $A_3$ or $S_3$; since $S_3$ is not abelian, $A\ne\{e\}$; since every member of $A$ is even, $H=A_3$.
\medskip
\quad(iii) $H$ must be either $\{e\}$ or the 4-element group $K$ or $A_4$. Since $A_4/K\cong\Bbb Z_3$ is abelian, $K$ includes $H$; since $A_4$ is not abelian, $H\ne\{e\}$; so $H=K$.
\medskip
(e)(i) Since $H\ne\{e\}$, $A$ must contain at least one 3-cycle; since the two 3-cycles are inverses of each other, it contains both; so $A=A_3$.
\medskip
\quad(ii) We know that $A$ contains at least one member of $K\setminus\{e\}$.
But since these all look exactly the same it had better contain all the others.
(This is a touch unrigorous, failing a proper definition of `look exactly the same'. Alternatively: all the members of $K\setminus\{e\}$ are conjugate in $A_4$, e.g.,
$\cycle{1 2 3}\cycle{1 2}\cycle{3 4}\cycle{3 2 1}=\cycle{2 3}\cycle{1 4}$, so if $A$ contains one of them it must contain the others.) So $A=K$.
}%end of ansCD
\def\qnDA{In the dihedral group $D_n$ of order $2n$, set
$\rho=\cycle{1 2\ldots n}$ and $\sigma=\cycle{1 n}\cycle{2\,\,n-1}\ldots$.
(a) Show that any subgroup of
$<\rho>=\{e,\rho,\rho^2,\ldots,\rho^{n-1}\}$ is a normal subgroup of $D_n$. ({\it Hint\/}: show that $(\sigma\rho^j)\rho^i(\sigma\rho^j)^{-1}=\rho^{-i}$.)
(b) Find $\rho\sigma\rho^{-1}$.
(c) Show that if $n$ is {\it odd}, then the only normal subgroup of $D_n$ containing $\sigma$ is $D_n$ itself. ({\it Hint\/}: $\rho=\rho^{n+1}$ is a power of $\rho^2$.)
(d) Show that if $n$ is {\it even}, then
\Centerline{$H^*=\{\sigma^i\rho^{2j}:0\le i<2,\,0\le j<\Bover{n}2\}$}
\noindent is a subgroup of $D_n$. Is it a normal subgroup? Why?
(e) Show that if $n$ is odd, then the only normal subgroups of $D_n$ are $D_n$ itself and the subgroups of $<\rho>$.
(f) Show that if $n$ is even, then the only normal subgroups of $D_n$ are $D_n$ itself, subgroups of $<\rho>$, and $H^*$.
(g) To which familiar group is the group $H^*$ of (d) above isomorphic?
(h) How many elements of $D_n$ are conjugate to $\rho$ in $D_n$?
How many elements of $D_n$ are conjugate to $\sigma$ in $D_n$? (Make sure your answer considers both odd and even $n$.)
\dbhfill[DA]}
\def\ansDAa{(a) If $H$ is a subgroup of $<\rho>$ it is certainly a subgroup of $D_n$. If $h\in H$ and $a\in D_n$, express $h$ as $\rho^i$ and $a$ as either $\rho^j$ or $\sigma\rho^j$. Then $aha^{-1}$ is either
\Centerline{$\rho^j\rho^i\rho^{-j}=\rho^i\in H$}
\noindent or
\Centerline{$\sigma\rho^j\rho^i\rho^{-j}\sigma^{-1}
=\sigma\rho^i\sigma^{-1}
=(\sigma\rho\sigma^{-1})^i=(\rho^{-1})^i=(\rho^i)^{-1}\in H$.}
\noindent So we always have $aha^{-1}\in H$; as $a$, $h$ are arbitrary, $H\normalsubgroup D_n$.
\medskip
{\bf (b)} $\rho\sigma\rho^{-1}=\sigma\rho^{-1}\rho^{-1}=\sigma\rho^{-2}$.
\medskip
{\bf (c)} If $\sigma\in H\normalsubgroup D_n$, then $H$ must contain $\sigma\rho^{-2}$ so contains $\rho^{-2}$ and $\rho^2$ and all its powers. But as $n$ is odd, $\rho=\rho^{n+1}$ is a power of $\rho^2$ and belongs to $H$.
So $H$ contains both $\rho$ and $\sigma$ and must be the whole of $D_n$.
\medskip
{\bf (d)} Because $n$ is even, $H^*=\{\sigma^i\rho^{2j}:i,\,j\in\Bbb Z\}$. Now
$e=\sigma^0\rho^0\in H^*$,
$$\eqalign{\rho^{2i}\cdot\rho^{2j}&=\rho^{2(i+j)}\cr
\sigma\rho^{2i}\cdot\rho^{2j}&=\sigma\rho^{2(i+j)}\cr
\rho^{2i}\cdot\sigma\rho^{2j}
&=\sigma(\sigma\rho^{2i}\sigma^{-1})\rho^{2j}
=\sigma\rho^{-2i}\rho^{2j}
=\sigma\rho^{2(j-i)}\cr
\sigma\rho^{2i}\cdot\sigma\rho^{2j}
&=\sigma^2\rho^{2(j-i)}
=\rho^{2(j-i)}\cr}$$
\noindent belong to $H^*$ for all $i$, $j\in\Bbb Z$, so $H^*$ is closed under multiplication. Finally
\Centerline{$(\rho^{2i})^{-1}=\rho^{-2i}$,\quad
$(\sigma\rho^{2i})^{-1}=\rho^{-2i}\sigma$}
\noindent belong to $H^*$ (the latter because it is the product of two members of $H^*$) for every $i\in\Bbb Z$, so $H^*$ is closed under inversion.
Thus $H^*$ is a subgroup of $D_n$. It is normal because it has just $n=\Bover12\#(D_n)$ members.
\medskip
{\bf (e)} Let $H$ be a normal subgroup of $D_n$ which is not a subgroup of $<\rho>$. Then $H$ contains $\sigma_1=\sigma\rho^i$ for some $i$. But now $H$ contains $\rho\sigma_1\rho^{-1}=\sigma_1\rho^{-2}$; so $H$ contains $\rho^2$ and $\rho$, and must be the whole of $D_n$.
\medskip
{\bf (f)} Let $H$ be a normal subgroup of $D_n$ which is not a subgroup of $<\rho>$. Then $H$ contains $\sigma_1=\sigma\rho^i$ for some $i$, and (as above) $\rho^2$ and all its powers belong to $H$. So either $H$ contains $\rho$, and is equal to $D_n$, or it contains $\sigma\rho^{2i}$ for every $i$, and includes $H^*$. In the latter case, because $\#(H^*)=\Bover12\#(D_n)$, $H$ is either $H^*$ or $D_n$.
\medskip
{\bf (g)} $D_{n/2}$.
}%end of ansDAa
\def\ansDAh{{\bf (h)} $\rho^i\rho(\rho^i)^{-1}=\rho$,
$(\sigma\rho^i)\rho(\sigma\rho^i)^{-1}=\rho^{-1}$, so just two elements of $D_n$ are conjugate to $\rho$ in $D_n$.
$\rho\sigma\rho^{-1}=\sigma\rho^{-2}$, so
$\rho^i\sigma(\rho^i)^{-1}=\sigma\rho^{-2i}$ for every $i\in\Bbb N$ (induce on $i$) and
\Centerline{$(\sigma\rho^i)\sigma(\sigma\rho^i)^{-1}
=\sigma(\rho^i\sigma\rho^{-i}\sigma^{-1}=\sigma\sigma\rho^{-2i}\sigma
=\sigma\rho^{2i}$}
\noindent for every $i\in\Bbb N$. Thus the elements of $D_n$ conjugate to $\sigma$ are of the form $\sigma(\rho^2)^i$ for some $i\in\Bbb Z$; if $n$ is even there are $n/2$ of them; if $n$ is odd there are $n$ of them.
(This corresponds to the fact that if $n$ is odd, all axes of reflection are of the same type, while if $n$ is even, they are of two types. $\sigma$, as defined here, is a reflection in an axis through the midpoint of an edge, and such reflections are conjugates of $\sigma$. If $n$ is even, then half the reflections are about axes through vertices, and these aren't conjugate to $\sigma$.)
}%end of ansDA
\def\qnDB{(a) Sketch a regular tetrahedron with the edges labelled 1 to 6. Write out, in disjoint-cycle form, permutations of the edges corresponding (i) to a rotation of the tetrahedron through an angle of $120^{\circ}$ about an axis through a vertex and the centre of the opposite face (ii) to a rotation of the tetrahedron through an angle of $180^{\circ}$ about an axis through the midpoints of two opposite edges (iii) to a reflection of the tetrahedron across a plane through an edge and the midpoint of the opposite edge.
(b) Show that $A_6$ has a subgroup $H$, isomorphic to $S_4$, which is {\it transitive}, that is, for every $i$, $j\le 6$ there is a $\rho\in H$ such that $\rho(i)=j$.
(c) Find a non-transitive subgroup of $A_6$ which is isomorphic to $S_4$.
\dbhfill[DB]}
\def\ansDB{(a)(i) $\cycle{123}\cycle{456}$
\quad(ii) $\cycle{25}\cycle{36}$
\quad(iii) $\cycle{26}\cycle{35}$.
(Of course these depend on the labelling of the vertices and on which rotations/reflections you pick. But any representation must give
double-3-cycles and double-transpositions as above.)
\medskip
(b) Because every simple exchange of vertices in the tetrahedron gives an even permutation of the edges, so does every rotation/reflection of the tetrahedron. So the action of the isometry group of the tetrahedron corresponds to a homomorphism from $S_4$ to $A_6$. Also any isometry which keeps all the edges in place must be the identity, so this homomorphism is injective and its image is a subgroup $H$ of $A_6$ isomorphic to $S_4$. Since any edge can be moved to any other by an isometry, $H$ is transitive.
\medskip
(c) I expect there are lots; but an easy one is just
\Centerline{$K=\{\rho:\rho\in A_6,\,\{\rho(5),\rho(6)\}=\{5,6\}\}$.}
\noindent (The point is that every member of $K$ permutes the numbers 1,2,3,4, and that for any permutation of 1,2,3,4 there is exactly one member of $K$ acting in that way, because if the permutation is even we need to set $\rho(5)=5$ and $\rho(6)=6$ to keep it even, and if it is odd we need to set $\rho(5)=6$ and $\rho(6)=5$.)
}%end of ansDB
\def\qnDC{(a) Show that the group of rotations of a cube is isomorphic to $S_4$.
(b) Give geometric descriptions of those rotations which correspond (i) to members of $K$, the four-element normal subgroup of $S_4$ (ii) to members of $A_4$, the alternating group.
(c) Give geometric descriptions of (i) the two types of rotation of order 2 (ii) the rotations of order 3 (iii) the rotations of order 4. How many of each of these are there?
\dbhfill[DC]}
\def\ansDC{(a) Consider the action of rotations of the cube on the four long diagonals. This corresponds to a homomorphism from the rotation group to $S_4$. If we look at a $180^{\circ}$ rotation about an axis through the midpoints of two opposite edges, this exchanges two of the diagonals and leaves the other two unmoved; so every transposition is in the image $H$ of the rotation group, and $H$ must be $S_4$. Since the number of rotations is $6\cdot 4=8\cdot 3=24=\#(S_4)$, the homomorphism must be injective as well as surjective and is an isomorphism.
[Alternative argument: A $90^{\circ}$ rotation about an axis through the midpoints of opposite faces corresponds to a 4-cycle in $S_4$, and a $120^{\circ}$ rotation about a long diagonal corresponds to a 3-cycle. So $H$ has subgroups of orders 3 and 4 and has at least 12 members. But the only subgroup of $S_4$ with 12 members is $A_4$, which does not contain any 4-cycle, so $H$ must be the whole of $S_4$.]
\medskip
(b)(i) $e\in K_4$ corresponds to the identity rotation. A
double-transposition in $K_4$ corresponds to a rotation through $180^{\circ}$ about an axis through the midpoints of two opposite faces.
The elements of $K_4$ correspond to rotations which do not move the three axes of the cube which pass through the midpoints of two opposite faces.
\medskip
\quad(ii) A 3-cycle corresponds to a rotation through $120^{\circ}$ about a long diagonal; other members of $K_4$ as above.
\medskip
(c)(i) We have already seen double-transpositions in (b-i); there are 3 of them, corresponding to the 3 pairs of opposite faces. As for simple transpositions, these are got by rotating $180^{\circ}$ about an axis through the midpoints of opposite edges; there are 6 of them, corresponding to the 6 pairs of opposite edges.
\medskip
\quad(ii) As noted in (b-ii), 3-cycles are rotations about long diagonals; there are 8 of them, 2 for each diagonal axis.
\medskip
\quad(iii) 4-cycles are $90^{\circ}$ rotations about axes through the midpoints of opposite faces; there are 6 of them, 2 for each axis.
}%end of ansDC
\def\qnDD{(a) Show that the group of rotations of a regular dodecahedron is isomorphic to $A_5$.
(b) Show that the group of isometries of a regular dodecahedron is not isomorphic to $S_5$.
(c) Give geometric descriptions of (i) the rotations of order 3 (ii) the rotations of order 5 (iii) the rotations of order 2. How many of each are there?
\dbhfill[DD]}
\def\ansDD{(a) We can inscribe 5 cubes in the surface of a regular dodecahedron; each cube has each of its 12 edges lying in a different face of the dodecahedron, being one of the 5 diagonals of that face. Two cubes meet at each vertex of the dodecahedron. Each isometry of the dodecahedron shuffles the five cubes; thus we have an action of the isometry group $G$ of the dodecahedron on the set of cubes, which can be interpreted as a homomorphism $\phi:G\to S_5$.
$\#(G)=60$. \Prf\ The dodecahedron has 12 faces, and each face can be oriented in 5 directions, so $\#(G)=5\cdot 12=60$.\ \Qed
$\phi[G]$ contains a 5-cycle. \Prf\ Let $a\in G$ be a rotation through an angle of $72^{\circ}$ about an axis through the centres of two opposite faces. Looking at the 5 diagonals of that face, each belonging to a different cube, we see that $\phi(a)$ is a 5-cycle.\ \Qed
$\phi[G]$ contains a 3-cycle. \Prf\ Let $b\in G$ be a rotation through an angle of $120^{\circ}$ about an axis through two opposite vertices $v$, $v'$.
Then $b^3=e$ so $\phi(b)$ is either the identity or has order 3. But if we look at any of the cubes not using the vertices $V$ and $V'$, we see that they are moved. So $\phi(b)$ is not the identity; as the only permutations in $S_5$ of order 3 are 3-cycles, $\phi(b)$ is a 3-cycle.\ \Qed
$\phi[G]$ contains a double-transposition. \Prf\ Let $c\in G$ be a rotation through $180^{\circ}$ about an axis through the midpoints of two opposite edges $E$ and $E'$. Then $c^2=e$ so $\phi(c)$ is either the identity, a transposition or a double-transposition. But the four cubes which use the vertices at the ends of $E$ and $E'$ are exchanged in pairs, so $\phi(c)$ moves 4 points and must be a
double-transposition.\ \Qed
Now consider $H=A_5\cap\phi[G]$. This is a subgroup of $A_5$ and contains elements of orders 2, 3 and 5. So $\#(H)$ is a multiple of
$2\cdot 3\cdot 5=30$. But $A_5$ is simple, so has no subgroup of order
30 (which would have to be normal). This means that $\#(H)=60$. This tells us that $\phi[G]\supseteq A_5$. But as $\#(G)=\#(A_5)=60$, $\phi[G]=A_5$ and $\phi$ is injective, therefore an isomorphism, and $G\cong A_5$.
\medskip
(b) Consider the isometry $t$ which takes each vertex to its opposite vertex. Then $t^2$ is the identity $\iota$. Also $tf=ft$ for every isometry $f$, because if $v$ is any vertex then $v$, $t(v)$ are opposite so $f(v)$, $f(t(v))$ are opposite, that is, $(t\smallcirc f)(v)=(f\smallcirc t)(v)$. But this means that $\{\iota,t\}$ is a normal subgroup of the isometry group. As $S_5$ has no normal subgroup of order 2, it cannot be isomorphic to the isometry group.
\medskip
(c)(i) A rotation of order 3 is a rotation through $120^{\circ}$ about a long diagonal. There are 20 of these (two for each of 10 diagonals).
\medskip
\quad(ii) A rotation of order 5 is a rotation through $72^{\circ}$ about an axis through the midpoints of opposite faces. There are 24 of these (four for each of 6 axes).
\medskip
\quad(iii) A rotation of order 2 is a rotation through $180^{\circ}$ about an axis through the midpoints of opposite edges. There are 15 of these (since the dodecahedron has 30 edges in 15 pairs).
}%end of ansDD
\def\qnDE{Consider a double-tetrahedron $X$ obtained by glueing two regular tetrahedra together at a common face. Let $G$ be the group of isometries of $X$, and $H$ the group of rotations.
(a) What are $\#(G)$ and $\#(H)$? Show that $H\normalsubgroup G$.
(b) How many elements of order 3 does $G$ have? How many of these are in $H$?
(c) How many elements of order 2 does $G$ have? How many of these are in $H$?
(d) Find an element of order 2 in the centre $C$ of $G$. ($C=\{a:ab=ba$ for every $b\in G\}$.) What is $C$?
(e) How many normal subgroups does $G$ have? Describe each normal subgroup of $G$ in geometric terms.
\dbhfill[DE]}
\def\ansDE{ I will call the three points of the glued-together faces the {\it equatorial points} of $X$ and the other two points the {\it poles}.
\medskip
(a) There are 6 rotations (2 places for a pole, then three for one of the equatorial points); 12 isometries (if the poles and an equatorial point are fixed, you can exchange the other two equatorial points). $H\normalsubgroup G$ because $\#(H)=\bover12\#(G)$.
\medskip
(b) $G$ has 2 elements of order 3, all of which are in $H$. (If $f\in G$ has order $3$, it must leave the poles unexchanged;
now it must act as a 3-cycle on the equatorial points.)
\medskip
(c) An element of order 2 in $G$ must either be a transposition of two points, which could be the poles or a pair of equatorial points (4 possibilities), or a double-transposition involving the poles and a pair of equatorial points (3 possibilities). So $G$ has 7 elements of order 2, of which 3 belong to $H$.
\medskip
(d) $C=\{\iota,t\}$, where $\iota$ is the identity rotation and $t$ is the reflection in the equator, exchanging the poles.
\medskip
(e) The normal subgroups of $G$ are
\quad$\{\iota\}$, the identity alone;
\quad$C=\{\iota,t\}$, the isometries which keep all equatorial points fixed;
\quad$H$, the group of rotations;
\quad the subgroup $S$, isomorphic to $S_3$, of the 6 isometries which leave the poles unexchanged;
\quad$S\cap H=\{\iota,g,g^2\}$ where $g$ is a $120^{\circ}$ rotation about an axis through the poles; the subgroup, isomorphic to
$\Bbb Z_3$, of rotations which leave the poles unexchanged.
}%end of ansDE
\def\qnEA{(a) Show that every element of $A_5$ is conjugate to its inverse in $A_5$.
(b) Show that every element of every $S_n$ is conjugate to its inverse in $S_n$.
(c) Show that for $n=3$, 4, 7 or 8 there is an element of $A_n$ which is not conjugate in $A_n$ to its inverse. What happens in $A_6$ and $A_9$?
\dbhfill[EA]}
\def\ansEA{(a) Look at the different possibilities.
\quad(i) If $\rho=e$ then $\rho^{-1}=eee^{-1}$ is conjugate to $\rho$.
\quad(ii) If $\rho=\cycle{i_1\,i_2\,i_3}$ then
$\rho^{-1}=\cycle{i_3\,i_2\,i_1}=\sigma\rho\sigma^{-1}$ where $\sigma=\cycle{i_1\,i_3}\cycle{i_4\,i_5}$.
\quad(iii) If $\rho=\cycle{i_1\,i_2}\cycle{i_3\,i_4}$ then
$\rho^{-1}=\rho=e\rho e^{-1}$.
\quad(iv) If $\rho=\cycle{i_1\,i_2\,i_3\,i_4\,i_5}$ then
$\rho^{-1}=\cycle{i_5\,i_4\,i_3\,i_2\,i_1}=\sigma\rho\sigma^{-1}$ where $\sigma=\cycle{i_1\,i_5}\cycle{i_2\,i_4}$.
\medskip
{\bf (b)} If $\rho\in S_n$, then $\rho^{-1}$ has a disjoint-cycle representation with exactly the same shape as that of $\rho$, so one can be converted into the other by some $\sigma\in S_n$.
\medskip
{\bf (c)} In $A_3$ and $A_4$, $\rho=\cycle{1 2 3}$ is not conjugate to its inverse $\cycle{3 2 1}$, because if $\sigma(1)=3$, $\sigma(2)=2$ and $\sigma(3)=1$ then $\sigma=\cycle{1 3}$ is odd. In $A_7$ and $A_8$, $\cycle{1 2 3 4 5 6 7}$ is not conjugate to its inverse.
In $A_6$, every permutation is conjugate to its inverse (calculations as in $A_5$, a little more complicated). In $A_5$,
$\cycle{1 2 3}\cycle{5 6 7 8}$ is not conjugate to its inverse.
\medskip
*If $\rho_1$, $\rho_2\in S_n$ have disjoint-cycle representations of the same form, they are conjugate in $S_n$, and conversely. If $\rho_1=\cycle{1\ldots k_1}\cycle{k_1+1\ldots k_1+k_2}\cycle{k_2+1\ldots k_2+k_3}\ldots$, then it is conjugate in $A_n$ to $\rho_2$ iff $\rho_2=\cycle{\sigma(1)\ldots\sigma(k_1)}
\cycle{\sigma(k_1+1)\ldots\sigma(k_1+k_2)}\ldots$ for an even permutation $\sigma$. Now if $\rho_1$ and $\rho_2$ are conjgate in $S_n$, we know that there is a permutation $\sigma$ which does the trick, but it need not be even.
If there is an {\it odd} permutation $\sigma_1$ such that $\sigma_1\rho_1\sigma_1^{-1}=\rho_1$, then one of $\sigma$, $\sigma_1$ will be even, and both will transform $\rho_1$ to $\rho_2$. So in this case $\rho_1$ will certainly be conjugate to $\rho_2$ in $A_n$. There will be such a $\sigma_1$ if {\it either} $\rho_1$ has a cycle of even length (we can take $\sigma_1$ to be that cycle) or $\rho_1$ has two cycles of the same odd length (we can take $\sigma_1$ to be the product of transpositions which exchanges the two cycles). If $\rho_1$ has only cycles of different odd lengths (including 1-cycles), then there is no such $\sigma_1$, because the only $\sigma$ such that $\sigma\rho_1\sigma^{-1}=\rho_1$ are products of powers of the cycles belonging to $\rho_1$.
In this second case, $\rho_2$ will be conjugate to $\rho_1$ in $A_n$ iff {\it any} $\sigma$ such that
\Centerline{$\rho_2=\cycle{\sigma(1)\ldots\sigma(k_1)}
\cycle{\sigma(k_1+1)\ldots\sigma(k_2)}\ldots$}
\noindent is even. So to determine whether $\rho_1$ and $\rho_1^{-1}$ are conjugate in $A_n$, we simply need to determine the parity of the permutation $\sigma$ which reverses all the cycles of $\rho_1$. Now if $\rho_1$ has cycles of odd lengths $k_1=2r_1+1,\ldots,k_m+2r_m+1$ then we need just $r_1+r_2+\ldots+r_m$ transpositions to turn them all round.
Thus $A_n$ will have an element $\rho$ not conjugate to its inverse in $A_n$ iff there are distinct numbers $r_1<\ldots1$. But $\#(\Orb(x))$ is a factor of $\#(G)=p^r$, so must be a multiple of $p$. This shows that $\#(G\setminus C)$ is a multiple of $p$ and $\#(C)$ is a multiple of $p$. In particular, $C\ne\{e\}$.
\medskip
{\bf (d)} Because $C\le G$, $\#(C)$ must be a factor of $p^2$, that is, $1$ or $p$ or $p^2$. By (c), it cannot be 1. By (b), $G/C$ cannot be of order $p$, so $\#(C)\ne p$. We are left with only one possibility, $\#(C)=p^2$, so that $C=G$ and $G$ is abelian.
\medskip
{\bf (e)} $D_4$ has $8=2^3$ members; if we take the standard generators $\rho$, $\sigma$ such that $\rho^4=\sigma^2=e$ and
$\sigma\rho\sigma^{-1}=\rho^{-1}$, the centre is $\{e,\rho^2\}$.
}%end of ansEB
\def\qnEC{(a) What is an {\it automorphism}? What is an {\it inner automorphism}? Explain why an inner automorphism, on the definition you give, is an automorphism. Show that the set $\Aut G$ of automorphisms of a group is again a group.
(b) Let $G$ be a group, $C$ its centre, $J$ the set of inner automorphisms of $G$. Show that
\quad(i) $J$ is a subgroup of $\Aut G$ isomorphic to $G/C$;
\quad(ii) $J\normalsubgroup\Aut G$.
(c) For each of the following groups $G$ describe $C$, $\Aut G$ and $J$:
(i) $G=K$, the four-element normal subgroup of $S_4$;
(ii) $G=S_3$;
(iii) $G=\Bbb Z_6$;
(iv) $G=\Bbb Z_5$;
(v) $G=D_4$, the dihedral group of order 8.
\dbhfill[EC]}
\def\ansECa{{\bf (a)} If $G$ is a group, an {\it automorphism} of $G$ is an isomorphism from $G$ to itself.
If $a\in G$, then the function $x\mapsto\hat a(x)=axa^{-1}:G\to G$ is an automorphism. \Prf\ $\hat a(xy)=axya^{-1}=axa^{-1}aya^{-1}$ for all $x$, $y\in G$, so $\hat a$ is a homomorphism. If $axa^{-1}=aya^{-1}$ then $x=y$ by the cancellation law, so $\hat a$ is injective. If
$y\in G$ then $\hat a(a^{-1}ya)=aa^{-1}yaa^{-1}=y$, so $\hat a$ is surjective, therefore an automorphism.\ \QeD\ Automorphisms of this kind are called inner automorphisms.
The identity function from $G$ to itself is an isomorphism, so belongs to $\Aut G$. If $\phi$, $\psi\in\Aut G$, then the composition $\phi\psi$ belongs to $\Aut G$, because the composition of two isomorphisms is an isomorphism. If $\phi\in\Aut G$, then
$\phi^{-1}\in\Aut G$, because the inverse of an isomorphism is an isomorphism. So $\Aut G$ is a subgroup of the group $S_G$ of bijections from $G$ to itself, and is a group in its own right.
\medskip
{\bf (b)(i)} Consider the conjugacy action of $G$ on itself given by the formula $a\action x=\hat a(x)=axa^{-1}$. We know that this corresponds to a homomorphism $a\mapsto\hat a:G\to S_G$. The image of this homomorphism is $J$, which must therefore be a subgroup of $S_G$; because also $J\subseteq\Aut G$, as noted in (a), $J$ is a subgroup of $\Aut G$. The kernel of the homomorphism is
\Centerline{$\{a:\hat a=\iota\}=\{a:axa^{-1}=x\Forall x\in G\}
=\{a:ax=xa\Forall x\in G\}=C$.}
\noindent So $C\normalsubgroup G$ and $G/C\cong J$, by the First Isomorphism Theorem.
\medskip
\quad{\bf (ii)} Now suppose that $a\in G$ and $\phi\in\Aut G$, and consider $\phi\hat a\phi^{-1}$. Set $b=\phi(a)$. Then for any $x\in G$,
$$\eqalign{(\phi\hat a\phi^{-1})(x)
&=\phi(\hat a(\phi^{-1}(x)))
=\phi(a\phi^{-1}(x)a^{-1})\cr
&=\phi(a)\phi(\phi^{-1}(x))\phi(a^{-1})
=bxb^{-1}=\hat b(x)\in J.\cr}$$
\noindent As $\phi$ and $a$ are arbitrary, $J\normalsubgroup\Aut G$.
}%end of ansECa
\def\ansECc{{\bf (c)(i)} As $G$ is abelian, $C=G$ and $J=\{\iota\}$. Since the multiplication table of $G$ is symmetric in the three elements of order 2 (the product of any two of them is the third), any permutation of these elements corresponds to an automorphism of $G$, and
$\Aut G\cong S_3$.
\medskip
\quad{\bf (ii)} $C=\{e\}$ so $J\cong G$. There are three elements of $G$ of order 2, and any element of $G$ is a product of these; so an automorphism of $G$ is determined by its values on the transpositions, and there can be at most $3!$ automorphisms. As $\#(J)=6$, $J=\Aut G$ and $\Aut G\cong S_3=G$.
\medskip
\quad{\bf (iii)} $C=G$ so $J=\{\iota\}$. $G$ has two elements of order 6; an element of $\Aut G$ must either keep them in their places or exchange them; since both generate $G$, $\Aut G$ has at most two members. Because $G$ is abelian, $x\mapsto x^{-1}$ is an automorphism, and $\Aut G\cong\Bbb Z_2$.
\medskip
\quad{\bf (iv)} $C=G$ so $J=\{\iota\}$. If $\phi:G\to G$ is an automorphism, set $j=\phi(1)$. Then $\phi(2)=j+_5j=2\times_5j$, $\phi(3)=j+_5j+_5j=3\times_5j$ etc., where $+_5$, $\times_5$ are addition and multiplication mod 5. Because $5$ is prime, this defines an automorphism of $G$ for any $j\in\Bbb Z_5^*=\{1,2,3,4\}$. The group operation of $\Aut G$ corresponds to $\times_5$ on $\Bbb Z_5^*$; so that $2\times_52=4$ and $2\times_52\times_52=3$, and
$\Aut G\cong(\Bbb Z_5^*,\times_5)\cong(\Bbb Z_4,+_4)$ is cyclic.
\medskip
\quad{\bf (v)} If we express $G=D_4$ in terms of generators $\rho$ and $\sigma$ where $\rho^4=\sigma^2=e$ and
$\sigma\rho\sigma^{-1}=\rho^{-1}$, then $C=\{e,\rho^2\}$. Since $x^2\in C$ for every $x\in G$, $J\cong G/C$ has every element of order 1 or 2, and is congruent to $K$.
Next, $\Aut G\cong D_4$. \Prf\ If $\phi\in\Aut G$, then $\phi(e)=e$ and $\phi[C]=C$, so $\phi(\rho^2)=\rho^2$. Let $Q=\{\sigma,\sigma\rho,\sigma\rho^2,\sigma\rho^3\}$ be the other four members of $G$ of order $2$. Then $\phi(a)\in Q$ for every $a\in Q$. What this means is that $\Aut G$ acts on $Q$. Because every element of $G$ can be expressed as a product of elements of $Q$, this action is faithful and $\Aut G$ can be identified with a subgroup of the group $S_Q$ of bijections from $Q$ to itself. If we arrange the elements of $Q$ in order around a square, we see that every element of $\Aut G$ leaves opposite corners as opposite corners (because it leaves $\rho^2$ unmoved), so $\Aut G$ corresponds to a subgroup of $D_4$. To see that it corresponds to the whole of $D_4$, observe that the defining relations remain true of $\rho$ and $\sigma\rho$ (since $(\sigma\rho)^2=e$ and $(\sigma\rho)\rho(\sigma\rho)^{-1}=\rho^{-1}$), so there is a $\psi_1\in\Aut G$ with $\psi_1(\rho)=\rho$, $\psi_1(\sigma)=\sigma\rho$, corresponding to a $90^{\circ}$ rotation of $Q$; and also that $\hat\sigma(\sigma\rho^j)=\sigma\rho^{-j}$, so that $\hat\sigma$ corresponds to a reflection in the diagonal $(\sigma,\sigma\rho^2)$. So $\Aut G$ corresponds to the whole of $D_4$.\ \Qed
}%end of ansEC
\def\qnEE{Let $G$ and $H$ be groups and $\phi:H\to\Aut G$ a homomorphism. Define an operation $*$ on $G\times H$ by writing
\Centerline{$(g_1,h_1)*(g_2,h_2)=(g_1\cdot\phi(h_1)(g_2),h_1h_2)$
for all $g_1$, $g_2\in G$, $h_1$, $h_2\in H$.}
(a) Show that $(G\times H,*)$ is a group.
(b) Show that $\{(g,e_H):g\in G\}$ is a normal subgroup of $G\times H$ isomorphic to $G$.
(c) Show that $(e_G,h):h\in H\}$ is a subgroup of $G\times H$ isomorphic to $H$.
(d) Show that if $Q$ is any group, $G\normalsubgroup Q$, $H\le Q$,
$G\cap H=\{e\}$ and $GH=Q$, then there is a homomorphism $\phi:H\to\Aut G$ such that $Q\cong(G\times H,*)$ if we define $*$ from $\phi$ as above.
($(G\times H,*)$ is called a {\it wreath product}.)
\dbhfill[EE]}
\def\ansEEa{{\bf (a)} If $g_1$, $g_2$, $g_3\in G$ and $h_1$, $h_2$, $h_3\in H$ then
$$\eqalign{(g_1,h_1)*((g_2,h_2)*(g_3,h_3))
&=(g_1,h_1)*(g_2\cdot\phi(h_2)(g_3),h_2h_3)\cr
&=(g_1\cdot\phi(h_1)(g_2\cdot\phi(h_2)(g_3)),h_1h_2h_3)\cr
&=(g_1\cdot\phi(h_1)(g_2)\cdot\phi(h_1)(\phi(h_2)(g_3))),h_1h_2h_3)\cr
&=(g_1\cdot\phi(h_1)(g_2)\cdot\phi(h_1)\circ\phi(h_2))(g_3)),
h_1h_2h_3)\cr
&=(g_1\cdot\phi(h_1)(g_2)\cdot\phi(h_1h_2)(g_3)),h_1h_2h_3)\cr
&=(g_1\cdot\phi(h_1)(g_2),h_1h_2)*(g_3,h_3)\cr
&=((g_1,h_1)*(g_2,h_2))*(g_3,h_3).\cr}$$
\noindent Thus $(G\times H,*)$ is a semigroup.
If $(g,h)\in G\times H$, then
\Centerline{$(g,h)*(e_G,e_H)=(g\cdot\phi(h)(e_G),he_H)=(ge_G,h)=(g,h)$.}
\noindent Set $g_1=\phi(h^{-1})(g^{-1})$, $h_1=h^{-1}$; then
$$\eqalign{(g_1,h_1)*(g,h)
&=(\phi(h^{-1})(g^{-1})\cdot\phi(h^{-1})(g),h^{-1}h)\cr
&=(\phi(h^{-1})(g^{-1}g),e_H)
=(\phi(h^{-1})(e_G),e_H)
=(e_G,e_H).\cr}$$
\noindent By the result in \number\qnnumberAA(c),
$(G\times H,*)$ is a group.
[Of course it is also easy to check directly that $(e_G,e_H)*(g,h)=(g,h)$ and $(g,h)*(g_1,h_1)=(e_G,e_H)$.]
\medskip
{\bf (b)} If $g_1$, $g_2\in G$ then
\Centerline{$(g_1,e_H)*(g_2,e_H)=(g_1\cdot\phi(e_H)(g_2),e_He_H)
=(g_1g_2,e_H)$.}
\noindent Thus the map $g\mapsto(g,e_H):G\to G\times H$ is a homomorphism. Since it is injective, its image $G'$ is a subgroup of $G\times H$ isomorphic to $G$.
If we look at the second-coordinate map $(g,h)\mapsto h:G\times H\to H$, we see that this is a homomorphism with kernel $G'$; so $G'$ is a normal subgroup of $G\times H$.
}%end of ansEEa
\def\ansEEc{{\bf (c)} If $h_1$, $h_2\in H$ then
\Centerline{$(e_G,h_1)*(e_G,h_2)=(e_G\cdot\phi(h_1)(e_G),h_1h_2)
=(e_Ge_G,h_1h_2)=(e_G,h_1h_2)$.}
\noindent Thus the map $h\mapsto(e_G,h):H\to G\times H$ is a homomorphism. Since it is injective, its image is a subgroup of $G\times H$ isomorphic to $H$.
\medskip
{\bf (d)} For $h\in H$, $g\in G$ set $\phi(h)(g)=hgh^{-1}$. Because $G\normalsubgroup Q$, $\phi(h):G\to G$ is a function; because the conjugacy action corresponds to a homomorphism from $H$ to $\Aut Q$, $\phi$ is a homomorphism from $H$ to $\Aut G$. Let * be the corresponding product on $G\times H$. Define $\psi:G\times H\to Q$ by setting $\psi(g,h)=gh$ for $g\in G$, $h\in H$. If $g_1$, $g_2\in G$ and $h_1$, $h_2\in H$ then
$$\eqalign{\psi((g_1,h_1)*(g_2,h_2))
&=\psi(g_1\cdot\phi(h_1)(g_2),h_1h_2)
=\psi(g_1h_1g_2h_1^{-1},h_1h_2)\cr
&=g_1h_1g_2h_1^{-1}h_1h_2
=g_1h_1g_2h_2
=\psi(g_1,h_1)\psi(g_2,h_2).\cr}$$
\noindent Thus $\psi$ is a homomorphism. Because $GH=Q$, $\psi$ is surjective. If $g\in G$, $h\in H$ are such that $\psi(g,h)=e$, then
$gh=e$, so $g=h^{-1}\in G\cap H$, $g=e$ and $h=e$; thus the kernel of $\psi$ is $\{(e,e)\}$, the trivial subgroup of $G\times H$, and $\psi$ is injective. So $\psi$ is an isomorphism.
}%end of ansEE
\def\qnEF{Let $G$ be a finite group and $H$ a subgroup of $G$.
(a) Taking $X$ to be the set of left cosets of $H$ in $G$, show that there is an action of $G$ on $X$ defined by setting $a\action D=aD$ for $a\in G$, $D\in X$.
(b) Let $K$ be the kernel of this action. Show that $K\subseteq H$. Show that $\Bover{\#(G)}{\#(K)}$ is a factor of $m!$, where $m=\Bover{\#(G)}{\#(H)}$.
(c) Show that if $K=H$ then $H\normalsubgroup G$.
(d) Show that if $m$ is a prime and is the smallest prime factor of $\#(G)$, then $H\normalsubgroup G$.
(e) Show that if $H$ is a proper subgroup of $A_5$, then $\#(H)\le 12$.
(f) For just which values of $n$ does $A_5$ have a subgroup of order $n$? Justify your answer.
\dbhfill[EF]}
\def\ansEF{{\bf (a)} If $D\in X$, then $D=dH$ for some $d\in G$, so $a\action D=a(dH)=(ad)H\in X$. Now if $a$, $b\in G$ and $D\in X$,
\Centerline{$a\action(b\action D)=a(bD)=(ab)D=ab\action D$,
\quad$e\action D=eD=D$.}
\medskip
{\bf (b)} Let $a\in K$. Since $H=eH\in X$, $aH=H$ so $a\in H$; thus $K\subseteq H$.
If we interpret the action as a homomorphism $\phi$ from $G$ to the group $S_X$ of bijections from $X$ to itself, $G/K\cong\phi[G]$, which is a subgroup of $S_X$. Now $\#(X)=\Bover{\#(G)}{\#(H)}=m$, by Lagrange's theorem, so $\#(S_X)=m!$ and $\#(G/K)=\#(\phi[G])$ is a factor of $m!$.
\medskip
{\bf (c)} Suppose that $K=H$ and that $a\in G$, $h\in H$. Then $haH=h\action(aH)=aH$, so $ha\in aH$. As $h$ is arbitrary, $Ha\subseteq aH$. Similarly, $Ha^{-1}\subseteq a^{-1}H$ so $aH\subseteq Ha$ and $aH=Ha$. As $a$ is arbitary,
$H\normalsubgroup G$.
\medskip
{\bf (d)} If $m$ is a prime and is the smallest prime factor of $\#(G)$, then $m$ is the greatest common divisor of $\#(G)$ and $m!$, so $\#(G/K)$ must be a factor of $m$, that is, $\Bover{\#(G)}{\#(K)}$ is a factor of $\Bover{\#(G)}{\#(H)}$. Since $K\subseteq H$, we must have $K=H$ and $H\normalsubgroup G$.
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{\bf (e)} Suppose that $H\le A_5$. Setting $m=\Bover{60}{\#(H)}$, $A_5$ has a normal subgroup $K$ such that $K\subseteq H$ and $\Bover{60}{\#(K)}$ is a factor of $m!$. But this means that either $K=\{e\}$ and $m\ge 5$ and $\#(H)\le 12$, or $K=A_5$ and $H=A_5$.
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{\bf (f)} $A_5$ has a subgroup of order $n$ iff $n=1$, 2, 3, 4, 6, 10, 12 or 60. \Prf\ By (e) and Lagrange's theorem, these are the only possibilities. For examples of subgroups of these orders:
\quad$\{e\}$ has order $1$.
\quad$<\cycle{1 2}\cycle{3 4}>$ has order $2$.
\quad$<\cycle{1 2 3}>$ has order $3$.
\quad$\{e,\cycle{1 2}\cycle{3 4},\cycle{1 3}\cycle{2 4},
\cycle{1 4}\cycle{2 3}\}$ has order $4$.
\quad$<\cycle{1 2 3 4 5}>$ has order $5$.
\quad Set $\rho=\cycle{1 2 3}$ and $\sigma=\cycle{1 2}\cycle{3 4}$. Then $\rho^3=\sigma^2=e$ and $\sigma\rho\sigma^{-1}=\rho^{-1}$. So $\{e,\rho,\rho^2,\sigma,\sigma\rho,\sigma\rho^2\}$ is a subgroup of $A_5$ of order 6.
\quad Set $\rho=\cycle{1 2 3 4 5}$ and $\sigma=\cycle{1 5}\cycle{2 4}$. Then $\rho^5=\sigma^2=e$ and $\sigma\rho\sigma^{-1}=\rho^{-1}$. So $\{e,\rho,\rho^2,\rho^3,\rho^4,\sigma,\sigma\rho,\sigma\rho^2,
\sigma\rho^3,\sigma\rho^4\}$ is a subgroup of $A_5$ of order 10.
\quad$A_4$ is a subgroup of $A_5$ of order 12.
\quad$A_5$ is a subgroup of $A_5$ of order 60.
}%\end of ansEF