# Proof that $\text{Aut}(G) \cong \text{GL}_{n}(p)$

I've been given this question as an extra task and I can't seem to make it work. Let $G=\underbrace{\mathbb{Z}/p\mathbb{Z}\times \ldots \times \mathbb{Z}/p\mathbb{Z}}_{n}$. Then I'm asked to prove that;

$$\text{Aut}(G) \cong \text{GL}_{n}(p)$$

This is just going to be my thought process so far on how to prove this;

Firstly, I noticed that $G= \{ (x_{1},\ldots,x_{n} \} \mid x_{i} \in \mathbb{Z}/p\mathbb{Z} \}$ ie the set of all $n$-tuples whoses enteries are in $\mathbb{Z}/p\mathbb{Z}$. As well as this, any automorphism of $G$ will take one $n$-tuple to another. This immediately hit me as being a linear transformation. Noting this, I know that this linear transformation could be represented by a matrix $M \in \text{GL}_{n}(p)$.Using this as inspiration, what I want to be able to do is create a homomorphism$$\psi : \text{Aut}(G) \to \text{GL}_{n}(p)$$

s.t $$\psi : \phi \to M$$

(ie it takes the mapping/linear transformation in $\text{Aut}(G)$ and represents it as a matrix.)

Here is where I get a little (a lot) stuck. I believe that showing it is a homomorphism is fairly straightforward, but showing that it is both on-to and one-to-one is evading me. If anyone could help me out I'd really appreciate it.

• The matrix of $\phi \in Aut(G)$ is obtained from $\phi(e_i)=\sum_{j=1}^n M_{i,j} e_j$ and the matrix of $\phi \circ \phi'$ is $M M'$. I'd say $GL_n(\mathbb{F}_p)$ is by definition the group of invertible $\mathbb{F}_p$-linear maps $\mathbb{F}_p^n \to \mathbb{F}_p^n$. If you meant $G$ is the additive group of $\mathbb{F}_p^n$, then $\phi \in Aut(G)$ is $\mathbb{Z}$-linear and hence $\mathbb{F}_p$-linear so that $GL_n(\mathbb{F}_p)= Aut(G)$. What you can show then is $GL_n(\mathbb{F}_p) = \{ M \in \mathbb{F}_p^{n \times n}, \det(M) \ne 0\}$. – reuns Oct 8 '17 at 18:42