3
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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 . . .

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  • 3
    $\begingroup$ If you don't want to use Sylow, can we use Cauchy? $\endgroup$ – Arthur Feb 18 at 13:18
  • $\begingroup$ No, @Arthur; it's not covered yet. Thank you though :) $\endgroup$ – Shaun Feb 18 at 13:21
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    $\begingroup$ If not then all nontrivial elements have order $7$, but $6$ does not divide $62$, so that is impossible. $\endgroup$ – Derek Holt Feb 18 at 13:25
  • $\begingroup$ That's very succinct, @DerekHolt. Thank you! :) $\endgroup$ – Shaun Feb 18 at 13:26

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