Call a group $G$ finitely generated if there is a finite subset $X \subseteq G$ with $G = \langle X \rangle$.
Prove that every subgroup $S$ of a finitely generated abelian group $G$ is itself finitely generated. (This can be false if $G$ is not abelian.)
Use induction on $n \geq 1$, where $X = \{a_1, \ldots ,a_n \}$. The inductive step should consider the quotient group $G/\langle a_{n+1} \rangle$.
Best Answer
We want to prove by induction on $n$ that every subgroup of an abelian group generated by $n$ elements is finitely-generated. If $n=1$, it is clearly true. From now on, suppose it is true for $n$.
Let $G= \langle a_1, \dots, a_{n+1} \rangle$ be an abelian group, $H \leq G$ be a subgroup and $\rho : G \to G/ \langle a_{n+1} \rangle$ be the quotient map. Notice that the group $G/ \langle a_{n+1} \rangle$ is generated by $\{ \rho(a_1) , \ldots, \rho(a_n) \}$, so by our induction hypothesis, the subgroup $\overline{H}:= \rho(H)$ is finitely generated: let $X=\{h_1,\dots, h_m\} \subset H$ be such that $\rho(X)$ generates $\overline{H}$.
On the other hand $H \cap \langle a_{n+1} \rangle$ is a cyclic group, say generated by $h_{m+1} \in H$. Now, we want to prove that $Y= \{h_1, \dots, h_m,h_{m+1} \}$ generates $H$.
Let $h \in H$. Of course, there exists a word $w \in \langle h_1,\dots, h_m \rangle$ such that $\rho(w)=\rho(h)$. Therefore, $h=w+k$ for some $k \in \mathrm{ker}(\rho)= \langle a_{n+1} \rangle$. Furthermore, $k=h-w \in H$ so $k=p \cdot h_{m+1}$ for some $p \in \mathbb{Z}$ since $\langle a_{n+1} \rangle \cap H = \langle h_{m+1} \rangle$. Finally, $$h=w+p \cdot h_{m+1} \in \langle h_1, \ldots, h_m,h_{m+1} \rangle = \langle Y \rangle,$$
so $Y$ generates $H$.