3
$\begingroup$

Let $G$ be a finite group.

I have to show that, for an automorphism $\alpha \in \text{Aut}(G)$, with no fixed points, which is defined as $$\{g \in G\mid\alpha(g) = g\} = \{e\},$$ with $e$ being the identity element of G, and further $\text{ord}(\alpha) = p$ for a prime $p$, then $p$ does not divide the group order $|G|$.

So far, I didn't really gain much from my trials of showing this, except that one basically has to prove that Cauchy's theorem doesn't hold, i.e. that, if $p$ does not divide $|G|$, then there does not exist any $g \in G$ with $\text{ord}(g) = p$, such that $p$ divides $|G|$. I thought one can do this with a proof by contradiction, i.e. assume that Chauchy's theorem holds, and then showing that such a $g \in G$ does not exist.

But I am not sure how to go about this, as I am confused about the "no fixed points" definition: if the set defined above is equal to $\{e\}$, and we have $\text{ord}(\alpha(g)) = \text{ord}(g) = p$, then how would one ever come to a contradiction here, since to me it seems as if the conditions already state that we have elements $g \in G$ with prime order.

$\endgroup$
2
  • $\begingroup$ I would look at a Sylow-$p$-subgroup of $\langle G,\alpha\rangle$ containing $\alpha$, and remember that the centre of a non-trivial $p$-group is non-trivial. $\endgroup$ Dec 9, 2022 at 13:47
  • 1
    $\begingroup$ You are overthinking this. Isn't it obvious that $|G| \equiv 1 \bmod p$? $\endgroup$
    – Derek Holt
    Dec 9, 2022 at 14:21

1 Answer 1

4
$\begingroup$

Hint: the $p$-group $\langle \alpha \rangle$ acts fixed point freely on $G$. Hence, $\{e\}$ is the only orbit of cardinality one and the rest of the orbits have length divisible by $p$ (in fact, since $\langle \alpha \rangle \cong C_p$, the others all have length equal to $p$) . It follows that $|G| \equiv 1$ mod $p$. So $|G|$ cannot be divisible by $p$.

$\endgroup$
2
  • $\begingroup$ Ah, thanks! I think I got it now. However, I don't see right now where the last conclusion, i.e. $|G| \equiv 1 \, \text{mod} \, p$ is coming from. Does this follow from the Sylow theorems? There's something I'm missing. $\endgroup$ Dec 11, 2022 at 17:23
  • 1
    $\begingroup$ Well the action of $\alpha$ splits $G$ up in a single orbit of length $1$ and the rest of the orbits of length $p$. No Sylow needed. Just plain set theory and counting. Perhaps you should study actions of groups on sets and the Orbit-Stabilizer theorem. $\endgroup$ Dec 11, 2022 at 22:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .