**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.11**

Solution: 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.