# Automorphism of a finite cylic $p$-group

Let $G$ be a finite group. Suppose that $A$ is a cylic $p$-subgroup and $A\unlhd N_G(Q)$ for some subgroup $Q$ of $G$.

Let $x \in N_G(Q)$ be an $p'$-order element. Then $x$ induces an automorphism of $A$ of order prime to $p$ and show that $x$ centralizes $A$

I know we can define an automorphism of $A$ by $a \mapsto xax^{-1}$ for $a\in A$. I'm not sure how the order is prime to $p$?

I also know that a cyclic group of order $p^n$, $n\geq 1$, always has an automorphism of order $p-1$, but I'm not sure how to use it to prove that $x$ centralizes $A$.

This is a proof in a journal paper by T.A Peng. I'm not sure which parts were relevant so I just included the entire initial assumptions and hypothesis.

• Why did you write "$x$ centralizes $A$" in the second paragraph? That does not follow. You are assuming that $x$ is a $p'$-element, so clearly the order of the automorphism induced is prime to $p$. Once again, many of your assumptions, like $A$ is a cyclic $p$-group, are irrelevant. – Derek Holt Jul 2 '16 at 14:24
• This is a proof in a journal paper by T.A Peng. I'm not sure which parts were relevant so I just included the entire initial assumptions and hypothesis. How do I determine the order of the induced automorphism? – R Maharaj Jul 2 '16 at 14:39
• If $x$ normalizes $A$, then the order of the induced automorphism is $|x|/|C_{\langle x \rangle}(A)$, which divides $|x|$. It doesn't follow from what you have written that $x$ centralizes $A$. – Derek Holt Jul 2 '16 at 14:42
• Now $|Aut(A)|$ is not divisible by any prime less than $p$ so how can we know that $x\in C_G(A)$? – R Maharaj Jul 2 '16 at 16:38
• That isn't true unless $p=2$. As you said yourself, $|{\rm Aut}(A)|$ is divisble by $p-1$. – Derek Holt Jul 2 '16 at 16:52

I found the paper, and I see that we are assuming at this point that $p$ is the smallest prime dividing $|G|$. If $|A|=p^n$, then $|{\rm Aut}(A)|=p^{n-1}(p-1)$, so the order of $x$ must be coprime to $|{\rm Aut}(A)|$ and hence the automorphism of $A$ that is induced by conjugation by $x$ is trivial. So $x$ centralizes $A$.

• Perfect! In future if I have any problems with statements of proofs in journal papers, is it alright to upload a screenshot of that part paper? – R Maharaj Jul 3 '16 at 2:24
• Final comment: if $x$ centralizes $A$ then $x$ centralizes $Q$ when $A\leq Q$? – R Maharaj Jul 3 '16 at 4:43
• If $x$ centralizes all cyclic subgroups of $Q$ then it centralizes $Q$. – Derek Holt Jul 3 '16 at 7:24