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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 $G/\langle a_{n+1} \rangle$.

user10024395
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1 Answers1

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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 $H$.

Seirios
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    can you explain your solution with more details? i don't quite understand your proof – user10024395 Nov 02 '14 at 18:19
  • I tried to be clearer, but please, be more precise when you need some explanation. – Seirios Nov 02 '14 at 19:15
  • why h = w + k ? and if k = h - w belongs to H, shouldn't k be generated by generators of H instead, why is it k = p h_m+1? – user10024395 Nov 03 '14 at 01:41
  • where is the condition of abelian used in your proof? – user10024395 Nov 03 '14 at 06:41
  • If $\rho(w)=\rho(h)$ then $\rho(w-h)=0$ ie. $w-h \in \mathrm{ker}(\rho)$. This means that $w=h+k$ for some $k \in \mathrm{ker}(\rho)$. Then, $k=p \cdot h_{m+1}$ for some $p \in \mathbb{Z}$ because I showed that $k \in H \cap \langle a_{n+1} \rangle = \langle h_{m+1} \rangle$. – Seirios Nov 03 '14 at 07:40
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    We used the fact that $G$ is abelian because we considered the quotient $G/ \langle a_{n+1} \rangle$ without wondering whether $\langle a_{n+1} \rangle$ is normal or not. – Seirios Nov 03 '14 at 07:41
  • so your $\rho$ map is a homomorphism? I think you need to give a definition of your map to make it more rigorous. How can we know if the map is homomorphic such that you can say p(w) = p(h) then p(w-h)? – user10024395 Nov 03 '14 at 07:49
  • Let us continue this discussion in chat. – Seirios Nov 03 '14 at 07:51
  • $\rho$ is the usual quotient map, that is if $g \in G$ then $\rho(g) \in G/ \langle a_{n+1} \rangle$ is the coset containing $g$. Indeed, it is a homomorphism. – Seirios Nov 03 '14 at 07:58