Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $G$ is a finite $p$-group (where $p$ is prime), so that $|G|=p^n$ for some positive integer $n\ge 2$.

How can we prove that $|\text{Aut}(G)|$ is divisible by $p$?

Here $\text{Aut}(G)$ is the group of all automorphisms of $G$. I know how to prove this when $G$ is non-abelian. In this case, we look at the action of group on itself by conjugation, i.e. we consider the map $\phi: G\to \text{Aut}(G)$ defined by $\phi(g)=\tau_g$ where $\tau_g: G\to G$ given by $\tau_g(x)=gxg^{-1}$ for each $x\in G$. The kernel of $\phi$ is then $Z(G)$, the center of the group. By First Isomorphism Theorem, we get that $G/Z(G)$ is isomorphically embedded in $\text{Aut}(G)$. Since $Z(G)$ is proper subgroup of $G$ (because $G$ is non-abelian), we see that $p$ divides $\left|G/Z(G)\right|$, so by Lagrange's Theorem, $\text{Aut}(G)$ is divisible by $p$.

But what happens when $G$ is abelian? The above homomorphism $\phi$ is no longer of use, since $\phi$ becomes the trivial map [i.e. $G=Z(G)=\text{ker}(\phi)$].

Thanks for your time :)

share|improve this question
    
I'm confused, the automorphism group of $\mathbb Z_p$ is not divisible by $p$ for any prime $p$. –  JSchlather Aug 21 '13 at 4:30
    
If $n=1$ this is not true. –  Mark Bennet Aug 21 '13 at 4:30
    
@JSchlather: Sorry about that! We need $n\ge 2$. –  Prism Aug 21 '13 at 4:32
    
The automorphism group of $\mathbb Z_2 \times \mathbb Z_2$ has order $3$. –  JSchlather Aug 21 '13 at 4:32
    
@Prism You're right, I recalled that it was $S_3$ and my group theory is rusty enough that I forgot about the factorial. At any rate, you just need to exhibit an automorphism of order $p$. I would argue using the fundamental theorem of finitely generated abelian groups. First demonstrate that $\mathbb Z_p \times \mathbb Z_p$ has an automorphism of order $p$. Show this reduces to the case that the group is cyclic and then show that $\varphi(p^n)$ is divisible by $p$ for $n>2$. –  JSchlather Aug 21 '13 at 4:37
show 5 more comments

1 Answer

up vote 4 down vote accepted

Use FTFGAG and consider two cases. (1) The group is elementary abelian and (2) it is not.

In the first case, the group is $\mathbb{Z}_p^n$. Its automorphism group clearly has order $(p^n-1)(p^n-p)\ldots(p^n-p^{n-1})$, which is divisible by $p$ if $n \geq 2$.

In the second case, the group is a direct sum of cyclic subgroups, and at least one of these subgroups has order greater than $p$. So $G \simeq \mathbb{Z}_{p^k} \oplus H$, where $k \geq 2$. $\mathrm{Aut} (G)$ then has a subgroup isomorphic to $\mathrm{Aut} (\mathbb{Z}_{p^k})$, and the latter has order $(p-1)p^{k-1}$, which is divisible by $p$.

share|improve this answer
    
Dan, thanks for your answer! Is it obvious that $\text{Aut}(K\oplus H)$ must have subgroup isomorphic to $\text{Aut}(K)$? I have the following idea: We can take all automorphisms of $K\oplus H$ that leaves $H$ fixed, and this should form a subgroup isomorphic to $\text{Aut}(K)$. Is my reasoning correct? –  Prism Aug 21 '13 at 4:54
1  
@Prism Almost. If $\varphi \in \mathrm{Aut}(K)$, then there's an automorphism of $K \oplus H$ that sends $(k, h)$ to $(\varphi(k), h)$ for any $k \in K, h \in H$. This gives an embedding of $\mathrm{Aut}(K)$ into $\mathrm{Aut}(K \oplus H)$, and the image leaves $H$ fixed. –  Dan Shved Aug 21 '13 at 5:06
    
Excellent! Thanks very much :) –  Prism Aug 21 '13 at 5:07
1  
@Prism Although, in the general setting, I see no reason why the image of $\mathrm{Aut}(K)$ must exhaust all the automorphisms of $K \oplus H$ that leave $H$ fixed. –  Dan Shved Aug 21 '13 at 5:10
    
Right. That's a subtle point. So it is possible that some automorphism of $K\oplus H$ leaves $H$ fixed, but when restricted to $K$, it is not an automorphism of $K$. But the other direction (given in your first comment) always works. –  Prism Aug 21 '13 at 5:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.