I want to prove that:

If $F$ is a free abelian group of rank $n$, then $\text{Aut}(F)$ is isomorphic to the multiplicative group of all $n\times n$ matrices over $\mathbb Z$ with $\text{det}=\pm1$.

What I have tried:

Since $F$ is a free abelian group of rank $n$ so I can write $F=\langle x_1,x_2,...,x_n\rangle$. If $\phi\in\text{Aut}(F)$ then it should map any $x_i$ to another element $x_j$ of the basis $X=\{x_1,x_2,...,x_n\}$. My idea is to map any element of $\text{Aut}(F)$ to a suitable matrix. For example when $n=2$: $$\phi_1:=\{x_1\to x_1, x_2\to x_2\}\Longrightarrow \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\\ \phi_2:=\{x_1\to x_2, x_2\to x_1\}\Longrightarrow \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$

Am I on a right approach? Or there is another better way? Thanks.

  • 1
    $\begingroup$ Word on notation, $M_n(\mathbb Z)$ usually denotes the abelian group [or the ring] of all $n\times n$ matrices. The multiplicative group of all invertible matrices is $GL_n(\mathbb Z)$. Note that in the case of $\mathbb Z$ a matrix is invertible if and only if its determinant is $\pm1$. $\endgroup$ – Asaf Karagila Oct 10 '12 at 14:06
  • $\begingroup$ @AsafKaragila: Thanks. I fixed it. $\endgroup$ – mrs Oct 10 '12 at 14:12
  • $\begingroup$ Show that $F\cong\Bbb Z^n$ and $\mathrm{End}(\Bbb Z^n)\cong M_n(\Bbb Z)$ (hence restricting each side to invertible elements, i.e. applying the "group of units" functor, you obtain $\mathrm{Aut}({\Bbb Z}^n)\cong\mathrm{GL}_n(\Bbb Z)$). You are going in the right direction, but why should each letter be sent to another letter? In general, automorphisms send generating sets to generating sets, but they do not necessarily preserve a given generating set, so for example with $n=2$, the unique map extending $x\mapsto2x+5y,y\mapsto x+3y$ will be an automorphism (note the underlying $\det$ is $1$). $\endgroup$ – anon Oct 10 '12 at 14:35

This sentence is not true: '$\phi$ should map any $x_i$ to another $x_j$', so it's not the best way.

But! We know the following:

  1. $\phi(x_i)=\sum_k \alpha_{ik}x_k$ for some $\alpha_{ik}\in\Bbb Z$, $k=1..n$.
  2. if $n\cdot w= \phi(x_i)$ for some $n\in\Bbb Z$ and $w\in F$, then $n\cdot\phi^{-1}(w)=x_i$, which is only possible for $n=\pm1$, as $x_i$ is an atom. But, for example $\phi(x_1)=x_1+x_2$ is possible..

So, 1. gives you the $A:=(\alpha_{ik})_{i,k}$ matrix. If you then write up the matrix $B$ of $\phi^{-1}$, you get that, $A,B\in\Bbb Z^{n\times n}$ and that $AB=BA=I$. So, $\det A$ is invertible and integer.

It remains to prove that all such matrices determine an automorphism of $F$. For that, if $\det A=\pm 1$, then it has an inverse over $\Bbb Q$, but using the Cramer form for the inverse, we get that all coefficients of $A^{-1}$ are actually integer (integer divided by $\det A$).

  • $\begingroup$ That was my clear mistake as @anon also noted. Thanks Belgi. $\endgroup$ – mrs Oct 10 '12 at 15:11

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