Show that $ℤ^{m}$ is a subgroup (and a free abelian group) of $ℤ^{n}$ for all $m≤n$

Question

My question is: Show that $ℤ^{m}$ is a subgroup (and a free abelian group) of $ℤ^{n}$ for all $0≤m≤n$.

Answer 1

If $m \le n$, then you can embed $\mathbb{Z}^m$ inside $\mathbb{Z}^n$ in the obvious way by the map $$(x_1,\ldots,x_m)\mapsto (x_1,\ldots,x_m,0,\ldots,0)$$ This is obviously an embedding (ie, an injective homomorphism)

Answer 2

Hint: What are possible (and simple) generators of $\mathbb{Z}^m$?