# Automorphism group of an abelian group

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.)

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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
-@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

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.