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\title{Group Theory: Math30038, Sheet 4}
\date{GCS}
\author{}
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\maketitle

\begin{enumerate}
\item {\em Suppose that $G$ acts on a set $\Omega.$ If
$\alpha \in \Omega$, we let
\[ G_\alpha = \{ g \in G \mid \alpha g = \alpha\}.\]
Now suppose that $\beta, \gamma \in \Omega$ are such that
$\beta h = \gamma$ for some $h \in G$. Show that
$G_\gamma = h^{-1} G_\beta h.$} \newline
\noindent {\bf Solution: } Suppose that $x \in G_\beta$,
then $\gamma h^{-1}xh = \beta h \cdot h^{-1}xh
= \beta xh = \beta h = \gamma$ so
$h^{-1} G_\beta h \subseteq G_\gamma$. Now $\beta =
\gamma h^{-1}$ so a similar argument shows that
$h G_\gamma h^{-1} \subseteq G_\beta$. Premultiplying
by $h^{-1}$ and postmultiplying by $h$ it follows that
$G_\gamma \subseteq h^{-1} G_\beta h.$ We have an inclusion
and its reverse, so $G_\gamma = h^{-1} G_\beta h.$ 
\item {\em Let $P$ be a group of order $p^n$ where $p$
is a prime number. Suppose that $P$ acts on a finite set $Q$
of size $q$ where $p$ does not divide $q$. Show that this action of $P$
has a fixed point (i.e. there is $\alpha \in Q$ such that
$\alpha g = \alpha \forall g \in P$).}  \newline  
\noindent {\bf Solution: } If $\xi \in Q$, then the stabilizer
(isotropy group) of $\xi$ is denoted $P_\xi$, and the orbit of $\xi$ has
size $|P:P_\xi|$ which is a power of $p$. Now count $Q$ by adding
up the sizes of the orbits of $P$ acting on $Q$ to discover
that at least one orbit must have size $1$, else $q$ would 
be divisible by $p$.


\item {\em In how many essentially different ways can one colour
the edges of a regular octahedron using $c$ colours (where
each edge is monochromatic, and two colourings are deemed the
same if one can moved to the other by a rigid motion -- and reflections
are not allowed).} \newline
\noindent {\bf Solution: } A regular octahedron has 8 
identical equilateral faces, and its group of rotational
symmetries $G$ has order 24. There are 4 axes of symmetry
through the centres of faces and the centre of the opposite face,
giving rise to 8 rotations $A$ of order 3. There are 3 axes of symmetry
through a vertex and the opposite vertex, giving rise to
6 rotations $B$ of order 4 and 3 rotations $C$ of order 2. There are 6 axes
of symmetry through the centre of an edge and the centre of the opposite
edge, giving rise to 6 rotations $D$ of order 2, and there is the identity map $E$.
We have described $8 + 6 + 3 + 6 + 1 = 24$ all of which are different,
and so have a list of all the elements of $G$. Now let $\Omega$
be the set of coloured octahedra, a set of size $c^{12}.$
We seek to count the orbits of $G$ acting on $\Omega$ by using
the counting principle not due to Burnside.
\[ \begin{array}{ccc}
\mbox{Element type} &  \mbox{Number of this type} &  |\mbox{Fix}| \\ 
      A      &          8     &      c^4\\
      B      &          6     &      c^2\\
      C      &          3     &      c^4\\
      D      &          6     &      c^4\\
      E      &          1     &      c^{12}
\end{array} \] 
The number of essentially different edge colourings of the octahedron
is therefore 
\[ \frac{c^{12} + 17c^4 + 6c^2}{24}.\] 
For example when $c = 2$ this is $183$. 
\item Let $G$ be a group with subgroups $H$ and $K$, each
of finite index in $G$. Prove that $H \cap K$ has finite index in
$G$. \newline 
\noindent {\bf Solution} It follows from Sheet 3, Question 4,
that $|K : H \cap K| \leq |G : H| < \infty.$ Now
$|G: H \cap K| = |G:K| \cdot |K: K \cap H| < \infty.$

\item {\em Let $G$ be a group and suppose that $H \leq G$ and
$|G : H | < \infty.$ By considering the groups $g^{-1}Hg$
as $g$ ranges over $G$ (or otherwise), prove that $G$ has
a normal subgroup $N$ with $|G : N | < \infty$ and $N \leq H \leq G$.}
\newline
\noindent {\bf Solution: } The solution to the previous question
shows that the intersection of two subgroups of finite index in $G$
is of finite index in $G$. A straightforward induction yields that
the intersection of finitely many subgroups of finite index in $G$ is
of finite index in $G$. Now, it is a routine matter to check (please
do it) that
each set $g^{-1}Hg$ is a subgroup of $G$ (where $g$ is an arbitrary element
of $G$). Let $\widehat H = \cap_{g \in G} g^{-1}Hg \leq G$. If
$x \in G$ and $y \in \widehat H$, then $x^{-1}yx \in (gx)^{-1} H gx$
for every $g \in G$ so $x^{-1}yx \in g^{-1}Hg$ for every $g \in G$.
Thus $x^{-1}\widehat H x \leq \widehat H$. Similarly
$x \widehat H x^{-1} \leq \widehat H$ and therefore
$\widehat H \leq x^{-1} \widehat H x$ for every $x \in G$ so
$\widehat H$ is a normal subgroup of $G$.

If $T$ is a right transversal
for $H$ in $G$, then it is easy to verify that $g^{-1}T$ is a right
transversal for $g^{-1}Hg$ in $G$ (please do it). Therefore each
group $g^{-1}Hg$ is of finite index in $G$. We will be finished if we
can show that there are only finitely many groups $g^{-1}Hg$ as
$g$ ranges over $G$. Suppose that $a, b \in G$ and that
$a^{-1}Ha \not = b^{-1}Hb$ so $(ab^{-1})^{-1}Hab^{-1} \not = H$
it follows that $ab^{-1} \not \in H$ and so $a$ and $b$ are in
different right cosets of $H$ in $G$. There are only finitely
many right cosets of $H$ in $G$ so there are only finitely many groups
$g^{-1}Hg$ as $g$ ranges over $G$.

\item {\em Let $G$ be a group and suppose that $x, y \in G$.
Prove that $o(xy) = o(yx).$ } \newline
\noindent {\bf Solution: }
$y (xy)^n = (yx)^n y$ so $(xy)^n = 1$ if and only if $(yx)^n =1$.
Thus the orders of $xy$ and $yx$ co-incide.
\end{enumerate}
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