If $G$ is a finite group and $\phi(x) = x^{p+1}$ is an automorphism of $G$ with $order(\phi) |p$ then $G$ is a $p$-group...?

If the order of $\phi$ is $1$ then $\phi(x) = x = x^{p+1} = x^px \rightarrow x^p =e$ so the order $\forall x\in G$ is $p$ therefore $G$ is a $p$-group

I'm having trouble when considering that the order of $\phi$ is $p$

If the order of $\phi$ is $p$ then

$\phi^p(x) =x= x^{(p+1)^p}$

using the binomial theorem, I get

$\forall x \in G \, \, \, order(x) | \ \displaystyle\sum_{k=1}^p \binom{p}{k}p^k = (p+1)^p-1$

At first, I thought that the only divisors of this was powers $p$ so I got this:

Suppose for contradiction that $G$ is not a $p$-group. Let $|G| = kp^n$ where $k$ is not a multiple of $p$ ($p\nmid k$). If $k>1$ then take a $q$ in $k$'s prime factorization. So we have $q|k$ and by Cauchy's Theorem $\exists y \in G$ with $y^q = e$ i.e $order(y) = q$.

Since $\forall x \in G \, \, \, order(x) | \ \displaystyle\sum_{k=1}^p \binom{p}{k}p^k$ we must have that $q|\displaystyle\sum_{k=1}^p \binom{p}{k}p^k$. A contradiction, therefore $k=1$ thus $G$ is a $p$-group.

But through examples, the divisors of $(p+1)^p-1$ also include other primes :'(. How can I show $G$ is a $p$-group?

Thanks so much :D


Taking the advice from the comments, Assume every proper subgroup of $G$ is a $p$-group, now take the Sylow $p$-subgroup of $G$, say $P$ then $P$ is normal since it is characteristic i.e $\phi(P) = P$. Since $P$ is a normal Sylow subgroup, it contains all $p$ subgroups of $G$ therefore $P$ contains all subgroups of $G$.

Can I conclude that $P=G$ since $P$ contains all subgroups of $G$?

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    $\begingroup$ Hint: Every subgroup of $G$ is $\phi$ invariant. You can assume by induction that every proper subgroup of $G$ is a $p$-group. $\endgroup$ – Geoff Robinson Jun 6 '14 at 18:15
  • $\begingroup$ @abe: that is called induction $\endgroup$ – Jack Schmidt Jun 6 '14 at 19:13
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    $\begingroup$ This question appears to be off-topic because it is obsolete. $\endgroup$ – user122283 Jun 7 '14 at 0:14
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    $\begingroup$ @SanathDevalapurkar There is a difference between off topic and obsolete! I don't think this question should be closed. $\endgroup$ – Pedro Tamaroff Jun 10 '14 at 3:07
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    $\begingroup$ There was no good reason, as far as I can tell, to close this question. $\endgroup$ – user98602 Jun 10 '14 at 3:17

This is false. Let $G$ be cyclic of order 7, and let $p=3$. Then $\phi:G \to G : g \mapsto g^4$ is an automorphism of order 3.

Update: The question has been changed now. You can read the original version here: https://math.stackexchange.com/revisions/823130/6

  • $\begingroup$ Thanks, I didn't notice it was wrong. -_- $\endgroup$ – abe Jun 6 '14 at 20:07
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    $\begingroup$ Jack: I know your a very knowledgeable person, but have you read the question carefully? The OP is asking that there is a nontrivial inner automorphism $axa^{-1}$ that acts like $x\mapsto x^{p+1}$. This smells like a semidirect product $(\Bbb Z_{p^2}\rtimes \Bbb Z_p$ using thr apropriate automorphism that gives the presentation $$\langle x,y\mid x^{p^2}=1,y^p=1,[y,x]=x^p\rangle$$ or something of the sort. Maybe the question is missing some hypotheses. $\endgroup$ – Pedro Tamaroff Jun 10 '14 at 3:17
  • $\begingroup$ @Pedro: the question has changed dramatically since my (correct) answer was posted. $\endgroup$ – Jack Schmidt Jun 10 '14 at 3:18
  • $\begingroup$ (For what is worth, I haven't read the question that carefully, but that is what I got from the scan!) $\endgroup$ – Pedro Tamaroff Jun 10 '14 at 3:18
  • $\begingroup$ @JackSchmidt Oh, my bad then. Let me see past edits. You can always call out on the OP to avoid these kind of misunderstandings! =) $\endgroup$ – Pedro Tamaroff Jun 10 '14 at 3:19

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