Is $p$ a prime divisor of $\phi(n)$ if $G$ has an element of order $n$?

[Euge]

Here is this week's POTW:

-----
Let $G$ be a noncyclic finite group of order $pn$ where $p$ is a prime such that $p!$ is coprime to $n$. Prove that if $G$ has an element of order $n$, then $p$ is a prime divisor of $\phi(n)$.

-----

 

[Euge]

The problem statement originally excluded the assumption that $G$ is noncyclic. The edit has been made. You can read my solution below.

Spoiler

Suppose $x\in G$ has order $n$. Let $G$ act on the set of left cosets of $\langle x\rangle$ by left multiplication. The permutation representation afforded by this action is a group homomorphism $G \to S_p$ whose kernel is $K = \cap_{g\in G} g\langle x\rangle g^{-1}$. By the first isomorphism theorem $(G : K)$ divides $|S_p| = p!$. Note $\langle x\rangle \supset K$ whence $|K| | n$ by Lagrange's theorem. So since $|G| = pn$ we have $\frac{n}{|K|} | \frac{p!}{p}$. Now $\frac{p!}{p}$ is coprime to $\frac{n}{|K|}$ since $p!$ is coprime to $n$; therefore $\frac{n}{|K|} = 1$, or $|K| = n$. Consequently $\langle x \rangle = K$ is normal in $G$.

By Cauchy's theorem, $G$ has an element $y$ of order $p$. Note $\langle y\rangle$ acts faithfully on $\langle x \rangle$ by conjugation. For if $\langle y\rangle$ acts trivially on $\langle x\rangle$, then $yxy^{-1}= x$, making $\langle x\rangle \cap \langle y \rangle = 1$. Then $G$ would be isomorphic $\langle x\rangle \times \langle y \rangle$, which is cyclic of order $pn$, contradicting the assumption on $G$ is noncyclic. So the action is nontrivial, which must be faithful since $\langle y\rangle$ has prime order. From the faithful representation $\langle y \rangle \to \operatorname{Aut}(\langle x\rangle) \cong \mathbb Z_n^\times$ we deduce $p | \phi(n)$, by the first isomorphism theorem.