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\title{Group Theory: Math30038, Sheet 5}
\date{Solutions GCS}
\author{}
\begin{document}
\maketitle

\begin{enumerate}
\item {\em 
Let $G = S_n$ be the symmetric group on $\{1,2,\ldots, n\}$, so
$|G| = n!$ and the elements of $G$ are the permutations of
$\{1,2,\ldots, n\}$.} \begin{enumerate}
\item {\em Suppose that $(a_1, a_2, \ldots , a_t) \in G$ is a cycle, and
that $g \in G$. Show that $g^{-1}(a_1, a_2, \ldots , a_t)g
= (a_1g, a_2g, \ldots , a_tg).$ } \newline
\noindent{\bf Solution: }\ If $x \not = a_i g$ for any $i$,
then $xg^{-1} \not = a_i$ for any $i$, so
$xg^{-1}(a_1, a_2, \ldots , a_t)g = xg^{-1}g = x$. However,
if $x = a_i g$ for some $i$, then 
$xg^{-1}(a_1, a_2, \ldots , a_t)g =  a_i g g^{-1}(a_1, a_2, \ldots , a_t)g \newline
= a_i (a_1, a_2, \ldots , a_t)g = a_{i+1}g$. Thus we can
describe \newline $g^{-1}(a_1, a_2, \ldots , a_t)g$ by the notation
$(a_1g, a_2g, \ldots , a_tg)$.
\item {\em Each element of $G$ can be expressed as a product of disjoint cycles
(elements of $G$ are {\rm disjoint} if their supports are disjoint). Show
that the number of conjugacy classes in $S_n$ is the number of
ways of writing $n$ as an ascending sum $a_1 + a_2 + \cdots + a_t$ of positive
integers $a_1 \leq a_2 \leq \cdots \leq a_t$. {\rm Thus there are 3
conjugacy classes in $S_3$ because 3 can be written as an ascending sum
in three ways: $3, 1+2, 1+ 1 +1.$ Also in $S_4$ there are 5 conjugacy classes be
cause
4 is $4, 1+3, 1+1 +2, 2+2$ and $1 + 1+ 1 + 1$.}} \newline
\noindent{\bf Solution: }\ Conjugation sends cycles to cycles
(of the same length). Morover, the calculation performed for Question
1 shows how $g$ must be chosen to effect the required
conjugation. In particular any two cycles of length $r$ are conjugate
in $S_n$ by a suitably chosen $g$. Conjugation sends disjoint
cycles to conjugate disjoint cycles so conjugacy classes in $S_n$ consist of all elements with a given `cycle shape'. Thus the cycle
shapes characterize the conjugacy classes, and these shapes can be
specified by listing the lengths of the disjoint cycles in
ascending order.
\item {\em Determine the number of conjugacy classes in $S_5$, and the size of each
conjugacy class, and describe the centralizer in $G$ of a chosen representative
of each conjugacy class.} \newline
\noindent{\bf Solution: }\ Here is a transversal for the conjugacy
classes: $id$, $(1,2)$, $(1,2,3)$, $(1,2,3,4)$, $(1,2,3,4,5)$,
$(1,2)(3,4,5)$ and $(1,2)(3,4)$. The corresponding conjugacy classes have
sizes $1, 10, 20, 30, 24, 20$ and $15$ and happily these numbers sum 
to $120 = 5!$. The corresponding centralizers must have orders
$120, 12, 6, 4, 5, 6$ and $8$. This enables us to identify them
as 
$G = S_5,$ 
$\langle (1,2),\ (3,4,5),\ (3,4)\rangle,$
$\langle (1,2,3),\ (4,5)\rangle,$  
$\langle (1,2,3,4)\rangle,$
$\langle  (1,2,3,4,5) \rangle,$ 
$\langle (1,2), (3,4,5) \rangle$
and $\langle (1,2,3,4), (1,2) \rangle.$
\end{enumerate}
\item {\em Let $G = S_n$. Let $x = (1,2,\ldots, n) \in G$. Prove that
$C_G(x) = \langle x \rangle.$} \newline
\noindent{\bf Solution: }\ The conjugacy class of $x$ has size $(n-1)!$
so $C_G(x)$ has index $(n-1)!$ and therefore order $n$.
However $\langle x \rangle \leq C_G(x)$ and $|\langle x \rangle| = n$
so $C_G(x) = \langle x \rangle.$
\item {\em Show that in $S_4$ there is a non-identity element $y$ such that
$C_G(y) \not = \langle y \rangle.$} \newline
\noindent{\bf Solution: }\ (1,2) centralizes (1,2)(3,4) (and vice versa).
\item {\em Suppose that $G$ is a finite group. Show that the number of elements
in each conjugacy class of $G$ must divide $G$.}  \newline
\noindent{\bf Solution: }\ Suppose that $x \in G$ and that $C$
is the conjugacy class of $x$. We know that $|C| = |G: C_G(x)|$
but $|G| = |G: C_G(x)| \cdot |C_G(x)|$ and we are done.
\item {\em Let $G$ be a group with a subgroup $H$ such that
$g^{-1}H g \subseteq H$ for every $g \in G$. Prove that
$g^{-1}H g = H$ for every $g \in G$.}  \newline
\noindent{\bf Solution: }\ This is a standard trick. If
$g^{-1}H g \subseteq H$ for every $g \in G$, then 
replacing $g$  by $g^{-1}$ we obtain that $gHg^{-1} \subseteq H$ for 
every $g \in G$. Therefore $H \subseteq g^{-1}H g$ for every $g \in G$
and so $H = g^{-1}H g$  for
every $g \in G$.
\item {\em (Challenge) Does there exist a group $G$ containing an element $g$
and a subgroup $H$ such that $g^{-1}Hg \subseteq H$ but $g^{-1}H g \not
= H$.}   \newline
\noindent{\bf Solution: }\ Yes. Let $G$ be the set of 2 by 2 invertible
rational matrices. Let $H$ be the set of upper unitriangular matrices
with integer entry in the top right position. Let $g$ be the diagonal matrix
$diag(1,2).$
\item {\em Suppose that $G$ is a group and that $H \leq G$. Choose $g \in G$.
Prove that $g^{-1}Hg \leq G$.} \newline
\noindent{\bf Solution: }\ $g^{-1}1g \in g^{-1}Hg \not = \emptyset.$
If $g^{-1}ag, g^{-1}bg  \in g^{-1}Hg$, then $g^{-1}agg^{-1}bg
= g^{-1}abg \in g^{-1}Hg$, As for inversion, the inverse of 
$g^{-1}ag$ is $g^{-1}a^{-1}g \in g^{-1}Hg$,
\item Let $G$ be a finite group of order $n$ which has $t$ conjugacy classes.
Elements $x$ and $y$ are each selected uniformly at random from $G$. What is
the probability that $x$ and $y$ commute? Does this make sense for abelian group
s? \newline
\noindent{\bf Solution: }\ Let $\Gamma = \{ (x, y) | x,y \in G, xy = yx \}.$
Now the required probability is
\[ p = \frac{1}{|G|^2} |\Gamma| = \frac{1}{|G|^2}\sum_{x\in G} |C_G(x)|\]
and by not Burnside's counting lemma this is $t/n$.
\item Show that a finite group with exactly two conjugacy classes must have two
elements.  \newline
\noindent{\bf Solution: } We have proved that the number of elements
in a conjugacy class of a finite group must divide the order of the group.
If the group order is $n$ this forces $n-1$ to divide $n$ which forces $n = 2$.
\item Let $G$ be a containing $H$ a subgroup of finite index. Let
$S = \{x^{-1}Hx \mid x \in G\}.$ Let $G$ act on $S$ by conjugation so
if $K \in S$ then $K \cdot g = g^{-1}Kg$. Verify that this is a group action,
and deduce that $|S| = |G:N_G(H)|$ where
$N_G(H) = \{ g \in G \mid gH = Hg \}.$ Deduce that $|S|$ is finite and
divides $|G:H|$.  \newline
\noindent{\bf Solution: } Suppose that $A \in S$ so that $A = x^{-1}Hx$
for some $x \in G$. Take elements $a,b \in G$. Now 
$A \cdot 1 = 1^{-1}A1 = A$ and $A \cdot ab = b^{-1}a^{-1}Aab
= (A \cdot a) \cdot b$. Thus we have a group action. There is a
single orbit of this action, and the orbit is in bijective correspondence
with the right cosets of the stabilizer of $H$. This is
$\{g \in G \mid g^{-1}Hg = H\}$ a group known as $N_G(H)$ and called
the normalizer of $H$ in $G$ (and of course $H \leq N_G(H)$ and in
fact $H \unlhd N_G(H)$). Thus $|S| = |G:N_G(H)|$ but
of course $|G:H| = |G:N_G(H)| \cdot |N_G(H):H|$  and we were
given that $|G:H| < \infty.$
Thus $|S|$ is finite and a divisor of $|G:H|.$
\end{enumerate}
\end{document}
