# For distinct primes $p$ and $q$, does any group of order $p^2q$ have a subgroup of order $p$ (without using Sylow or Cauchy)?

## 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.

$$\dagger$$ Yes, by Sylow's Theorems, right? But wait . . .
• If not then all nontrivial elements have order $7$, but $6$ does not divide $62$, so that is impossible. – Derek Holt Feb 18 at 13:25