# Groups acting on themselves

I am working on ex. 4.2.11 in Dummit and Foote's Abstract Algebra book. The question reads:

Let $G$ be a finite group and let $\pi : G\rightarrow S_{G}$ be the left regular representation. Prove that if $x$ is an element of $G$ of order $n$ and |$G$|=$mn$, then $\pi (x)$ is a product of $m$ $n$-cycles. Deduce that $\pi (x)$ is an odd permutation if and only if |$x$| is even and $\frac{|G|}{|x|}$ is odd.

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Let the subgroup generated by $x$ act on $G$ by left multiplication. –  Pierre-Yves Gaillard Oct 12 '11 at 3:08

By definition, $\pi(x): G \rightarrow G$ maps $g \mapsto xg$. Since the order of $x$ is $n$, for every $g \in G$, composing $\pi(x)$ $n$ times, we have $\pi(x) \circ \cdots \circ \pi(x) = x^n g = g$, and this is the smallest number of times we can compose $\pi(x)$ to get the identity map. So the orbit size of each $g$, i.e. the size of each disjoint cycle must be $n$, and since $|G| = mn$, there must be $m$ such cycles.
Each $n$-cycle can be written as the product of $n-1$ transpositions. Now $\pi(x)$ is an odd permutation iff $m(n-1)$ is odd iff $n = |x|$ is even and $m = |G|/|x|$ is odd.