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QUESTION

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\item[(a)]
State Burnside's lemma, explaining carefully any notation that you
use.

\item[(b)]
Let $G$ be a finite group of order divisible by 3.

\begin{description}

\item[(i)]
Let $X=\{(g,h,k)|g,h,k\in G,ghk=e\}$. Find the number of elements
of $X$.

\item[(ii)]
The cyclic group $\left<t\right>$ of order 3 acts on the set
$G\times G\times G$ via the rule $t(g,h,k)=(h,k,g)$. Show that
this defines an action of $\left<t\right>$ on $X$.

\item[(iii)]
Show that the fixed points of $t$ are precisely the elements of
order 3 in $G$.

\item[(iv)]
Apply Burnside's lemma to the action of $\left<t\right>$ on $X$ to
show that 3 must divide the number of fixed points for $t$, and
deduce that $G$ must have at least one element of order 3.

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ANSWER


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\item[(a)]
$$r|G|=\sum_{g\in G}|X_g|$$

where $G$ a group acts on a set $X$, $r$=number of orbits,
$|G|$=number of element in $G$ and for each $g\in G,\ X_g=\{x\in
X|gx=x\}$

\item[(b)]

\begin{description}

\item[(i)]
For any $g,h\in G$ there is a unique $k\in G$ with $ghk=e$, so
there are $|G|^2$ elements in $X$.

\item[(ii)]
It suffices to show that for any $(g,h,k)\in X\ (h,k,g)\in X$ too,
i.e. that $ghk=e\Leftarrow hkg=e$. But
$hkg=g^{-1}(ghk)g=g^{-1}eg=e$.

\item[(iii)]
$t(g,h,k)=(g,h,k)\Leftrightarrow(h,k,g)=(g,h,k)\Leftrightarrow
h=g=k$, so the fixed points for $t$ in $X$ are precisely the
triples $(g,g,g)$ such that $ggg=e$; i.e. $X_t=\{g\in G|g^3=e\}$

\item[(iv)]
$|X_t|=|X_{t^{-1}}|=|X_{t^2}|=$ number of elements of order 3 plus
1 (the identity).

So $r|\left<G\right>|=|X_e|+|X_t|+|X_{t^2}|=|X|+2|X_t|=|G|^2+2$
(number of elements of order 3 +1)

So 2(number of elements of order 3 +1) is divisible by
$|\left<t\right>|=3$. Hence number of elements of order 3$\neq0$.

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