Why is $\mathbb{Z}\otimes Ag = Ag$, where Ag is an abelian group

Question

Why is $$\mathbb{Z}\otimes Ag = Ag,$$ where Ag is an abelian group? In other words, why is it an identity? Perhaps an explanation of the simpler example of the $\mathbb{Z}\otimes \mathbb{Z}$ is equal to $\mathbb{Z}$.

I thank you in advance.

Answer 1

I believe the result is actually that $\mathbb{Z}\otimes Ag\cong Ag$. To prove this, we just need to define two homomorphisms $f\colon\mathbb{Z}\otimes Ag\to Ag$ and $h\colon Ag\to\mathbb{Z}\otimes Ag$ which are inverses of one another.

Well, then let's consider:

$$f(n,g)=ng\quad\text{and}\quad h(g)=(1,g)$$

First notice that $f$ is bilinear so it's well-defined. It's clear that $f(h(g))=g$. Furthermore $$g(f(n,g))=(1,ng)=(n,g)$$ so these are indeed inverses, which proves the statement.

Your statement that $\mathbb{Z}\otimes\mathbb{Z}\cong\mathbb{Z}$ obviously still holds as a particular case, I just didn't think it simplified matters.

I hope this helps.