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.
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.