Prove that $G$ is a free abelian group.

Let $G$ denote the set of all 4x4 matrices of the form

$$\begin{equation*} M(a,b) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ a & 1 & 0 & 0 \\ 0 & 0 & 1 & b \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{equation*} $$

I have already proven that $G$ is an abelian group under the ordinary multiplication of matrices. What I am having trouble with is proving that $G$ is a free abelian group. Basically I have to come up with a basis for $G$, but I am unsure if that implies a row basis, a column basis, or something else that I am not even considering. This was always very vague what this meant when it was introduced to me. Someone mind helping me sort this out so I can write a proof?

The other thing I have to do is determine the rank of the free abelian group, which I think will be straight forward if I can find that basis for proving $G$ is a free abelian group.

Solution 1:

I'm assuming $a, b \in \mathbb{Z}$, as you want this group to be free abelian. As is pointed out in the comments, if $R$ is a commutative ring, then the set of matrices of your form where $a, b \in R$ is isomorphic to $R^2$ as an $R$-module. That is, $G \cong R^2$ is the "free $R$-module on $2$ generators", and in the special case $R = \mathbb{Z}$, we see that $G \cong \mathbb{Z}^2$ is the "free abelian group on $2$ generators. Let's talk about how to prove this:

We see that matrices of $G$ are block diagonal, so it decomposes as a sum $G \cong G_1 \oplus G_2$ where $G_1$ are matrices of the form $\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$ and $G_2$ are matrices of the form $\begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix}$. Of course, it's "well known" that

$$ \begin{pmatrix} 1 & 0 \\ a_1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ a_2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ a_1 + a_2 & 1 \end{pmatrix} $$

So we see that $G_1 \cong (R,+)$ as groups. A similar computation (which it sounds like you've already done) shows that $G_2 \cong (R,+)$ too! In particular, this means

$$G \cong G_1 \oplus G_2 \cong R \oplus R \cong R^2.$$

Now, let's say that we're not confident with block diagonal decompositions. How can we use the idea of the previous argument to show directly that $G \cong \mathbb{Z}^2$?

Well, inspired by the above idea, we might expect $M(1,0)$ and $M(0,1)$ to be the two generators of our group (do you see why?). With this in mind, we can show directly that

$$ \begin{align} f &: \mathbb{Z}^2 \longrightarrow G \\ f &: (a,b) \mapsto M(a,b) \end{align} $$

is an isomorphism. After all, it's obviously injective and surjective (do you see why?) so all that remains is to check that it's a group homomorphism. Of course, I'll leave this computation to you!

I hope this helps ^_^