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I was going through the following post where it has been used that $Aut(\mathbb Z_3^2)\cong GL_2(\mathbb Z^2)$. My question is: why so? Is there ny general result on behalf of this ?

If so please provide me some link on it so that I can study them thoroughly

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KON3
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  • Ah. I suspect the original post is in error, especially since it makes no sense! It states $Aut(\mathbb{Z}^2_3)=GL_2(\mathbb{Z}^3)$, which is nonstandard. I suspect he means $Aut(\mathbb{Z}^2_3)=GL_2(\mathbb{Z}_3)$, which is almost a tautology. – Pax Sep 15 '15 at 08:55
  • The general result is $Aut \left( (\mathbb{Z}/m\mathbb{Z})^n \right) \cong GL_n(\mathbb{Z}/m\mathbb{Z})$. – Matt B Sep 15 '15 at 10:44

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This is obviously a typo: $GL_2(\mathbb Z^3)$ and $GL_2(\mathbb Z^2)$ are infinite, while a finite set like $\mathbb Z_3^2$ only has finitely many bijections from itself to itself.

What is meant is that $\operatorname{Aut}(\mathbb Z_3^2) \cong GL_2(\mathbb Z_3)$. This is just the usual statement from linear algebra that automorphisms of a vector space can be identified with invertible matrices, after a choice of basis. Here the vector space is $\mathbb Z_3^2$ over the field $\mathbb Z_3$.

The only thing you have to think about is that indeed every group automorphism of $\mathbb Z_3^2$ is also a $\mathbb Z_3$-vector space automorphism.

moonlight
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