Let $G$ be a finite group. If there exists an $a\in G$ not equal to the identity such that for all $x\in G$,$\phi(x) = axa^{-1}=x^{p+1} $ is an automorphism of $G$ then $G$ is a $p$-group.

This is what I have so for.

The order of $a$ is $p$ since $\phi(a) = a= a^pa\rightarrow a^p=e$ therefore the $order(\phi)|p$

If $order(\phi) = 1$ then for all $ x\in G$ $\phi(x) = x=x^{p+1}\rightarrow x^p=e$ . Thus every element has order $p$ therefore $G$ is a $p-group$

For $order(\phi) = p$ I get stuck:

$\phi^p(x) = x = x^{(p+1)^p}$ using the expansion formula and simplifying i reach that the order of each element in $G$ divides $\displaystyle\sum_{k=1}^n \binom{p}{k}p^k=(p+1)^p-1$. But other primes divide this so I can't easily conclude $G$ is a $p$-group.

I proceeded by contradiction:

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 $order(y) = q$.

If $q<p$, applying $\phi$ to $y$ I get that $\langle y\rangle$ has more than $q$ elements since $\phi(y)=y^{p+1}\in \langle y \rangle$ and there are $p$ distint elements achieved by $\phi$

If $p>q$ then....... I cannot reach a contradiction :'(

Question #2: Show each element of $G$ has order $p$

When $order(\phi) = 1$ i get what I want but I'm also stuck when $order(\phi) =p$

I believe $a\in Z(G)$, is there a way I can show this?

Thank you....This is a pretty hard problem. :'(

  • $\begingroup$ An earlier version of the problem did not mention $\theta$ is an inner-automorphism: math.stackexchange.com/questions/823130/… $\endgroup$ – Jack Schmidt Jun 10 '14 at 2:40
  • $\begingroup$ Sorry about I thought $axa^{-1}$ was just there to show the order of the automorphism is a divisor of $p$ $\endgroup$ – abe Jun 10 '14 at 2:48
  • $\begingroup$ It is not a good practice to use quantifiers such as $\forall$ and $\exists$ when writing mathematics. It makes the question harder to read. See math.harvard.edu/graduate/advise/kleiman.pdf $\endgroup$ – Mauricio Tec Jun 10 '14 at 3:04
  • $\begingroup$ :O What! But I took a semester learning this $\endgroup$ – abe Jun 10 '14 at 3:05
  • $\begingroup$ Should I not be using $\wedge$ $\neg$...either? $\endgroup$ – abe Jun 10 '14 at 3:06

Here is a plan.

Let $N = \{x \in G \mid x^p = 1\}$.

  1. Note that $x \in N$ if and only if $x$ commutes with $a$. It follows that $N$ is a subgroup of $G$. It is easy to see that $N$ is characteristic and therefore normal.

  2. Take an arbitrary $y \in G$. We have $a y a^{-1} = y^{p+1}$ in $G$. Project this equality onto $G/N$, it gives us $yN = y^{p+1}N$ ($a$ is gone because $a \in N$). Therefore, $y^p \in N$, therefore $y^{p^2} = 1$. This shows that $G$ is a $p$-group, even that the order of each element divides $p^2$.

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Here is a different proof from Dan Shved's:

Lemma: In any such group $G$, $a$ normalizes every subgroup, and $a$ lies in every Sylow $p$-subgroup.

Proof: Let $H$ be a subgroup. Then for $x \in H$, $axa^{-1}=x^{p+1} \in H$, so $aHa^{-1} \leq H$; a similar statement holds for $a^{-1}$ and so $H$ is normal. Let $P$ be a Sylow $p$-subgroup. Since $a$ has order $p$ and $a$ normalizes $P$, $\langle a, P \rangle$ is a $p$-group, and so by maximality of $P$, $\langle a, P \rangle = P$ and $a \in P$. $\square$

Main proof: Reduce to a minimal counterexample:

Let $G$ be a group of smallest order such that $G$ has an element $a$ of order $p$ such that for every $x \in G$, $ax=x^{p+1}a$ and yet $G$ is not a $p$-group. If $H$ is a subgroup of $G$ containing $a$, then $H$ satisfies the same hypothesis. If $H$ is also a proper subgroup, then $H$ is a $p$-group by definition of $G$.

Now consider what this means for an element of prime order $q\neq p$.

Let $g$ be an element of order $q$ for $q \neq p$. By hypothesis, $a$ normalizes $\langle g \rangle$ so $H=\langle a,g\rangle$ has order $pq$ and satisfies the hypothesis of the problem. If $H <G$ then we have a contradiction, so we must $H=G$ and so $G$ is a non-abelian group of order $pq$.

And show that this structure doesn't satisfy the hypothesis:

Let $P=\langle a \rangle$ be a Sylow $p$-subgroup of $G$ and let $Q=\langle g \rangle$ be a Sylow $q$-subgroup. Since $Q$ is normalized by both $P$ and $Q$, $Q$ is normal in $G$. Since $a$ is contained in every Sylow $p$-subgroup by the lemma, but the Sylow $p$-subgroups have order $p$, we have $P$ is also normal. Hence $G = P \times Q$ is abelian. This a contradiction, since then $aga^{-1} aa^{-1}g = g \neq g^{p+1}$ as $g$ has order $q$ not $p$.


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  • $\begingroup$ (This proof was based on a higher tech proof: $a$ lies in the Baer kernel of $G$, so is contained in the second center, so centralizes Sylow $q$-subgroups, a contradiction. I tried to get the conclusions using earlier ideas.) $\endgroup$ – Jack Schmidt Jun 10 '14 at 4:53
  • $\begingroup$ Thank you so much :D . I wish I can accept two answers. Sorry about yesterday, i must have driven you nuts. $\endgroup$ – abe Jun 10 '14 at 16:42

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