Caracterization of $\mathrm{Aut}(\mathbb{Z}^n)$ for $n \in \mathbb{N}$

Question

As explained in the title, I'm trying to characterize the group of automorphism of $\mathbb{Z}^n$ seen as a group.

Can anyone help me or work with me on this question?

P.B.

Is it isomorphic to semiproduct of other small groups? How?

Answer 1

Hint: $\mathbb{Z}^n$ is a free abelian group. Every homomorphism from $\mathbb{Z}^n$ is determined by the image of a basis. Every endomorphism is given by a matrix with integer entries.

Answer 2

This is the general linear group $GL_n(\mathbb{Z})$ consisting of matrices with integer coefficients such that there is an inverse with again integer coefficients. For a commutative ring, we know that $GL_n(R)=\{ A \in M_n(R)\mid \det(A)\in R^{\times}\}$. In other words, $$ {\rm Aut}(\mathbb{Z}^n)=GL_n(\mathbb{Z})=\{ A \in M_n(\mathbb{Z})\mid \det(A)=\pm 1 \}. $$ This is an arithmetic group. In particular, it has a finite presentation, and only finitely many conjugacy classes of finite subgroups.