Let $B_0=\{(2,2),(-4,4)\}$. Let $A={\rm span}_\Bbb Z(B_0)$. Prove that $A$ is a free abelian group of rank $2$.

Question

Let $B_0=\{(2,2),(-4,4)\}$. Let $A={\rm span}_\Bbb Z(B_0)$. Prove that $A$ is a free abelian group of rank $2$.

I know that A will cover all of the even numbers. However, I am not sure how to prove it is a free abelian group.

Is it enough to show that $B_0$ is a basis of A? Or would it be better to show that it is isomorphic to $\Bbb Z^2$?

For the rank I believe that it is just the cardinality of the basis (which would be B right? and since it has 2 elements it will have rank 2?

Answer 1

You need to show, first, that $B_0$ is linearly independent over $\Bbb Z$ before claiming that its cardinality is enough to show that $A$ has rank two.

It suffices, then, to show that, if $a=(2,2), b=(-4,4)$, then

$$A\cong\langle a,b\mid ab=ba\rangle;$$

that is, indeed, that $A\cong \Bbb Z^2.$