Showing that GL(F_p) is isomorphic to an automorphism group

[Mr Davis 97]

Homework Statement

Let $p$ be a prime integer. Show that $\operatorname{Aut}(\underbrace{Z_p\times \dots \times Z_p}_{n \text{ factors }})\cong GL_n(\mathbb{F}_p)$, where $\mathbb{F}_p$ is $Z_p$ viewed as a field.

Homework Equations

The Attempt at a Solution

First, note that $Z_p = \{1,x,\dots,x^{p-1}\}$.= is the cyclic group of order $p$. Also, let $e_i$ be the element in $(Z_p)^n$ with $x$ in the $i$-th position and the identity everywhere else.

Here is my idea of what an isomorphism would look like:

$\Psi : GL_n(\mathbb{F}_p) \to \operatorname{Aut}((Z_p)^n)$ given by $(v_1~\dots~v_n) \mapsto \alpha$ where $\alpha(e_i) = v_i$.

So here are my questions. How can I show that this map is well-defined in the sense that the $\alpha$ being mapped to is actually an automorphism? Also, how would I show that this map is onto?

 

[fresh_42]

Also, how would I show that this map is onto?

I see a fundamental difficulty here. To show all elements of $\operatorname{Aut}(\mathbb{Z}_p^n)$ are hit, you will have to know what they are. But if you know what they are, i.e. found a way to describe them, you probably will also have found a way to establish a bijection to $GL_n(\mathbb{F}_p)$. So your idea looks to me as a circle argument: Solve the problem in order to be able to solve the problem.

Another approach would be an induction along $n$. The situation $n=1$ and $n=2$ should already point out the way, i.e. how to translate $\operatorname{Aut}(\mathbb{Z}_p)$ to $\mathbb{F}_p^*=GL_1(\mathbb{F}_p)$ and how to deal with cross mappings, that is the anti-diagonal ($n=2$).