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.

We know that for the elementary abelian group $G$ of order $p^n$, its automorphism group is $GL(n,\mathbb{Z}_p)$. Here we consider the group $G$ as a vector space over the field $\mathbb{F}_p\cong \mathbb{Z}_p$; and every automorphism is an invertible linear transformation of the vector space and conversely, so the automorphism group is $GL(V)\cong GL(n,\mathbb{Z}_p)$.

Is it true that the automorphism group of the abelian group $(\mathbb{Z}_{p^2})^n$ is $GL(n,p^2)$? (I think that it need not be true; I could not apply the technique of above group, because here the group $\mathbb{Z}_{p^2}$ is a ring, but not a field, so I can not consider the group as a vector space.)

share|improve this question
    
WHat do you mean by $GL(n,p^2)$? That usually means the group of $n\times n$ matrices in a field of $p^2$ elements... –  Mariano Suárez-Alvarez Apr 22 '11 at 4:39
1  
-@Mariano: Oh! I have to specify it!! I was thinking $GL(n,p^2)$ as you written, the general linear group over the field of order $p^2$. –  user8186 Apr 22 '11 at 4:43

2 Answers 2

up vote 5 down vote accepted

If G is a finite direct product of cyclic groups of order pn, then Aut(G) is isomorphic to the group GL(n,Z/pnZ) of invertible n×n matrices over the ring Z/pnZ. These matrices work surprisingly similarly to matrices over a field. They are invertible if and only if their determinant is invertible (so not divisible by p).

G is called a homocyclic group. Aut(G) has a normal subgroup consisting of those matrices which act trivially on the vector space G/Gp. This normal subgroup is called a congruence subgroup and consists of those matrices that are congruent to the identity mod p.

General finite abelian groups have slightly more complicated automorphism groups, which are still matrix groups, but where the entries come from different rings and bi-modules (that is, there are different sorts of congruence conditions required). Similar techniques work for understanding finite-dimensional algebras.

share|improve this answer

You're right - it's not true (why would it be, really?). To see that it's enough to consider a single counterexample - just take $p=n=2$. The group $\mbox{Aut}(\mathbb{Z}_4 \times \mathbb{Z}_4)$ is of order 96, while $\mbox{GL}(2,4)$ is of order 180.

share|improve this answer

Your Answer

 
discard

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