Munkres' Section 67 Exercise 6 ( Rank of the subgroups of free abelian group with finite rank)

Question

Hi. I am studying this topic and I cannot see the statement "If this subgroups is nontrivial, choose $x_m\in B_m$ so that $\pi(x_m)$ is a generator of this subgroup.

I want proves that $\langle \pi(x_m)\rangle=\pi_m(B_m)$. Let $y\in \pi_m(B_m)$ then $y=\pi_m(x)$ some $x\in B_m$ but this $x$ depens of $y$...

Why exists that $x_m$?

Answer 1

$\Bbb Z$ is cyclic. So all its subgroups are cyclic.