Exercise 4.2.11 (Cycle decomposition of $\pi(x)$, where $\pi$ is the left regular representation)

Proof. By Section 4.2, we know that the regular representation is transitive and faithful, and so $\pi :G\to S_G$, defined by $\pi (x)(g)=\mathit{𝑥𝑔}$ for all $g\in G$, is injective. Therefore $\pi$ induces an isomorphism between $G$ and a subgroup $H$ of $S_G\simeq S_{\mathit{𝑚𝑛}}$.

Since isomorphisms preserve the order of the elements, for all $x\in G$

Put $\sigma =\pi (x)={\pi }_x$, and consider a cycle $\tau$ of $\sigma$ containing $a\in G$, so $\tau$ is given by

where the length $l$ of $\tau$ is the least positive integer such that ${\sigma }^{l}a=a$. Then, for all integers $k$,

This shows that $l=n$, so all cycles for $\sigma ={\pi }_x$ have the same length $n$: if

is the cycle decomposition of $\pi (x)$, then all cycles ${\tau }_i$ are $n$-cycles, where $n=|x|$ (see examples in Exercise 4.2.2, 4.2.3 and 4.2.4).

If $x=1$, then $\pi (x)=()$, and every $1$-cycle is the identity.

If $x\ne 1$, then $x\cdot g\ne g$ for all $g$, so no element of $G$ is fixed by $\pi (x)$.

This shows that no element $a\in G$ is fixed by $\pi (x)$ if $x\ne 1$, so the support of $\pi (x)$ is the whole set $G$, which is the union of the supports of the disjoint cycles ${\tau }_i$. This shows that $|G|=\mathit{𝑚𝑛}=\mathit{𝑘𝑛}$, so $k=m$.

In conclusion, if $|G|=\mathit{𝑚𝑛}$, where $n=|x|$, then $\pi (x)$ is a product of $m$ $n$-cycles.

Hence

Therefore