## Characterization of parity in the left regular representation of a finite group

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

This action of $G$ is faithful, so that the induced action of $H = \langle x \rangle$ on $G$ is faithful. Note that under this action, for all $g \in G$ we have $\mathsf{stab}_H(g) = 1$, so that $[H : \mathsf{stab}_H(g)] = n$. Thus each $H$-orbit of $G$ has order $n$; the $H$-orbit of an element $g \in G$ is precisely the cycle containing $g$ in the decomposition of $\pi(x)$ since $H$ is cyclic with generator $x$. Because $|G| = mn$, there are $m$ distinct orbits. Thus $\pi(x)$ is a product of $m$ disjoint $n$-cycles.

Clearly if $|x|$ is even and $|G|/|x|$ is odd, then $\pi(x)$ is an odd permutation by Proposition 25.

Now if $\pi(x)$ is an odd permutation and $|x|$ is odd, then the number of cycles in $\pi(x)$ of even length is zero, which is even, so that $\pi(x)$ is an even permutation- a contradiction. Thus $|x|$ is even. Now by Proposition 25, $|G|/|x|$ must be odd.

• Gobi Ree  On December 7, 2011 at 9:23 pm

Faithfulness does not imply that the stabilizer of any point is $1$. So the reasoning in the first paragraph is a little misleading. It will be better to say that:

This action of $G$ satisfies that the stabilizer of any point is the identity subgroup, so does the induced action.

• nbloomf  On December 9, 2011 at 1:43 am

Thank you; that was poorly worded.

• Gobi Ree  On December 7, 2011 at 9:33 pm

typo: $\mathsf{stab}_H(a)$$\mathsf{stab}_H(g)$
the cycle containing $G$ – the cycle containing $g$

• nbloomf  On December 9, 2011 at 1:07 am

Thanks!