# If $\alpha$ is an automorphism with no fixed points and $\text{ord}(\alpha) = p$, then the prime $p$ does not divide the group order

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.

• 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. Dec 9, 2022 at 13:47
• You are overthinking this. Isn't it obvious that $|G| \equiv 1 \bmod p$? Dec 9, 2022 at 14:21

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 $$p$$.
• 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. Dec 11, 2022 at 17:23
• 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. Dec 11, 2022 at 22:26