Show that there is no subgroup B of $\mathbb Z ^2$ such that $\mathbb Z ^2 / B \cong \mathbb Z_3 \times \mathbb Z_6 \times \mathbb Z_{21}$

Question

I have the following question in group theory:

Show that there is no subgroup $B$ of $\mathbb Z ^2$ such that $\mathbb Z ^2 / B \cong \mathbb Z_3 \times \mathbb Z_6 \times \mathbb Z_{21}$

I don't really know how to approach this question. I thought maybe looking at the generators on the left side, using the first isomorphism theorem, or looking at the prime decomposition, but I'm not really sure.
I could use a hint in this one. Group theory is not my strongest subject.

Answer 1

Any group isomorphic to some $\mathbb{Z}^2/B$ is commutative, and can be generated by two elements, namely the classes of $(1,0)$ and $(0,1)$.

Actually, this is an exact characterization: if $A$ is an abelian group generated by two elements $x$ and $y$, then we can define a surjective group morphism $f:\mathbb{Z}^2\to A$ by $f(1,0)=x$ and $f(0,1)=y$, and then if $B:= \ker(f)$, this gives $A\simeq \mathbb{Z}^2/B$.

Now obviously $\mathbb Z_3 \times \mathbb Z_6 \times \mathbb Z_{21}$ is commutative, so can it be generated by two elements?