The Details:
I'm reading "Contemporary Abstract Algebra (Eighth Edition)," by Gallian.
This is based on Exercise 7.40 of the "Cosets and Lagrange's Theorem" section ibid. Here it is for convenience:
Prove that a group of order $63$ must have an element of order $3$.
What I've noticed about the motivating question:
We have $63=3^2\times 7$ and we're looking for an element (and therefore a subgroup) of order $3$.
Everything I could find online or think of uses Sylow's Theorems. That would be an anachronism since they're not covered in the book so far, so, therefore, I suppose, there ought to be a proof sans this result. The same is true of Cauchy's Theorem.
But I'm thinking bigger . . . Hence:
The Question:
For distinct primes $p$ and $q$, does any group $G$ of order $p^2q$ have a subgroup of order $p$?$^\dagger$ Moreover, can this be proven without using Sylow's Theorems or Cauchy's Theorem?
Thoughts:
If $G$ is cyclic, then for any nontrivial $g\in G$, we have
\begin{align} e&=g^{|G|} \\ &= g^{p^2q} \\ &= (g^{pq})^p, \end{align}
so $g^{pq}$ has order (at most?) $p$ (but $p$ is prime so . . . ). Thus we are done.
I'm not sure what to do next. I have a vague idea of applying Cayley's Theorem followed by the Orbit-Stabiliser Theorem, although the former wouldn't tell us much about the orbit of an element.
Please help :)
$\dagger$ Yes, by Sylow's Theorems, right? But wait . . .