Want to prove $Aut(A_{n})\simeq GL(n, \mathbb{Z})$ where $A_{n}$ is a free abelian group of rank $n$

Question

I'm trying to prove that $Aut(A_{n}) \simeq GL(n, \mathbb{Z})$, where $A_{n}$ is a free abelian group of finite rank $n$.

I have already available to me the result that $A_{n} \simeq \mathbb{Z}^{n}$, so essentially what I have to prove is that $Aut(\mathbb{Z}^{n}) \simeq GL(n, \mathbb{Z})$.

To that effect, I considered $E = \{e_{1}, e_{2}, \cdots, e_{n}\}$ a basis of $\mathbb{Z}^{n} \simeq A_{n}$, where $\forall i$, $e_{i}$ is of the form $(0, 0, \cdots, 0, 1, 0, \cdots, 0)$ (all entries from $1$ to $i-1$ are $0$, the $i$th entry is $1$, and all entries from $i+1$ to $n$ are $0$.

Then, I let $\varphi: \mathbb{Z}^{n} \to \mathbb{Z}^{n}$ be the automorphism where $\varphi: e_{i} \mapsto \sum_{j=1}^{n}m_{ij}e_{j}$. (So, essentially, the $m_{ij}$ are "coordinates" of each $e_{i}$ in terms of the basis elements $e_{j}$).

Then, I thought about setting up a map $\mu: \varphi \mapsto M_{\varphi}$ where $M_{\varphi}$ would be an $n\times n$ matrix with integer entries $m_{ij}$.

Before I can even show that this map $\mu$ is an isomorphism, however, I must first show that it is a homomorphism.

So, to that effect, I thought I'd set up another automorphism $\psi: \mathbb{Z}^{n} \to \mathbb{Z}^{n}$ where $\psi: e_{i} \mapsto \sum_{j=1}^{n}k_{ij}e_{j}$. Then, I'd need to show that $\mu(\varphi \circ \psi) = \mu(\varphi(\psi(e_{i}))) = M_{\varphi}M_{\psi}$.

I'm not even sure that I defined $\psi$ correctly, and then when I tried to actually expand $\mu(\varphi \circ \psi)$ hoping I could get the matrices to pop out, I got as far as the following:

$\mu(\varphi \circ \psi) = \mu(\varphi(\psi(e_{i}))) = \mu(\varphi(k_{i1}e_{1} + k_{i2}e_{2} + \cdots + k_{ii}e_{i}+\cdots + k_{in}e_{n}) = \mu(\varphi(k_{i1}e_{1})+ \varphi(k_{i2}e_{2}) + \cdots + \varphi(k_{ii}e_{i})+ \cdots + \varphi (k_{in}e_{n})) = \mu\left( \sum_{j=1}^{n}k_{i1}m_{ij}e_{j} + \sum_{j=1}^{n} k_{i2}m_{ij}e_{j} + \cdots + \sum_{j=1}^{n} k_{ii}m_{ij}e_{j} + \cdots + \sum_{j=1}^{n}k_{in}m_{ij}e_{j}\right) = \mu \left(\left[k_{i1}m_{i1}e_{1} + k_{i1}m_{i2}e_{2} + \cdots + k_{i1}m_{ii}e_{i} + \cdots + k_{i1}m_{in}e_{n} \right] + \left[k_{i2}m_{i1}e_{1} + k_{i2}m_{i2}e_{2} + \cdots + k_{i2}m_{ii}e_{i} + \cdots + k_{i2}m_{in}e_{n} \right] + \cdots + \left[k_{in}m_{i1}e_{1} + k_{in}m_{i2}e_{2} + \cdots + k_{in}m_{ii}e_{i} + \cdots + k_{in}m_{in}e_{n} \right] \right)$

Then, I figure I'd need to collect terms in such a way as to, once applying $\mu$, I would obtain $M_{\varphi}M_{\psi} = \begin{pmatrix} m_{11} & m_{12} & \cdots & m_{1n} \\ m_{21} & m_{22} & \cdots & m_{2n} \\ \vdots & \vdots& & \vdots \\ m_{n1} & m_{n2} & \cdots & m_{nn} \end{pmatrix} \begin{pmatrix} k_{11} & k_{12} & \cdots & k_{1n} \\ k_{21} & k_{22} & \cdots & k_{2n} \\ \vdots & \vdots& & \vdots \\ k_{n1} & k_{n2} & \cdots & k_{nn}\end{pmatrix} = \begin{pmatrix} \sum_{i=1}^{n}m_{1i}k_{i1} & \sum_{i=1}^{n}m_{1i}k_{i2} & \cdots & \sum_{i=1}^{n}m_{1i}k_{in} \\\sum_{i=1}^{n}m_{2i}k_{i1} & \sum_{i=1}^{n}m_{2i}k_{i2} & \cdots & \sum_{i=1}^{n}m_{2i}k_{in} \\ \vdots & \vdots & & \vdots \\ \sum_{i=1}^{n}m_{ni}k_{i1} & \sum_{i=1}^{n}m_{ni}k_{i2} & \cdots & \sum_{i=1}^{n}m_{ni}k_{in} \end{pmatrix}$.

With what I've done so far is this even going to happen?

For the $\mu(\varphi(k_{i1}e_{1})+ \varphi(k_{i2}e_{2}) + \cdots + \varphi(k_{ii}e_{i})+ \cdots + \varphi (k_{in}e_{n})) = \mu\left( \sum_{j=1}^{n}k_{i1}m_{ij}e_{j} + \sum_{j=1}^{n} k_{i2}m_{ij}e_{j} + \cdots + \sum_{j=1}^{n} k_{ii}m_{ij}e_{j} + \cdots + \sum_{j=1}^{n}k_{in}m_{ij}e_{j}\right)$

part, can I pull the $k_{i1}, k_{i2}$, etc. out of the $\sum$'s and hence out of the $\varphi$'s?

In any case, if someone could please help me show that this map $\mu$ is a homomorphism, I'd greatly appreciate it. Thank you! I'm getting bogged down in these summations, and I'm figuring there has to be an easier way to show this.

Answer 1

Indeed, there's an easier way.

It may serve you well to pick up a linear algebra textbook and see how this is done for matrix-representations of linear maps on $\Bbb F^n$. Then again, there are some textbooks that also get bogged down in summations, so that depends on the textbook.

In any case, the quicker approach is as follows: to show that $\mu$ is a homomorphism, it suffices to note that $$ \mu(\varphi)v = \varphi(v) $$ for any $\varphi \in GL_n(\Bbb Z)$ and $v \in \Bbb Z^n$. Here, $\mu(\varphi)v$ denotes matrix multiplication, whereas $\varphi(v)$ is an application. With that, we can deduce that for any $v \in \Bbb Z^n$, $$ \mu(\varphi \circ \psi)v = (\varphi \circ \psi)(v) = \varphi(\psi(v)) = \mu(\varphi)\psi(v) = \mu(\varphi)\mu(\psi)v $$ from which we may conclude that $\mu(\varphi \circ \psi) = \mu(\varphi)\mu(\psi)$, as desired.