# If an abelian group $G$ has property $N$, then $G$ is finitely generated

An abelian group $G$ has property $N$ if for every sequence of subgroups $G_1 \subseteq G_2 \subseteq G_3 \subseteq \ldots \subseteq G$ there exists an integer $N$ such that $G_M=G_N$ for all $M \geq N$. Show that if $G$ has property $N$ then $G$ is finitely generated.

Here's what I did: Since for every element $\alpha \in G$, the subgroup $\langle\alpha\rangle$ is isomorphic to $\mathbb{Z}$ or $\mathbb{Z}/ n \mathbb{Z}$, we can arrange a sequence of subgroups $H_1 \subseteq H_2 \subseteq H_3 \subseteq \ldots \subseteq G$ where each $H_i$ is generated by some $\alpha_i$ in $G$. Since $G$ has property $N$, the sequence stabilizes at some point $H_N$. But I am unsure how to find a finite set of generators for $G$. Any hints?

• Use \langle and \rangle to produce $\langle$ and $\rangle$ as opposed to $<$ and $>$. – Michael Albanese Nov 16 '15 at 0:23

## 2 Answers

You almost have the right idea, but you don't want each of the $H_i$'s to be generated by a single element. In general, you can't order such subgroups by inclusion. Instead, you want the number of generators in each subgroup to be increasing.

Suppose that $G$ is not finitely generated. Given a non-identity element $\alpha_1 \in G$, $\langle\alpha_1\rangle \neq G$ as $G$ is not finitely generated; define $H_1 = \langle\alpha_1\rangle$. As $H_1 \neq G$, there is a non-identity element $\alpha_2 \in G\setminus H_1$. Again, as $G$ is not finitely generated, $\langle\alpha_1, \alpha_2\rangle \neq G$. Define $H_2 = \langle\alpha_1, \alpha_2\rangle$; note that $H_1 \subseteq H_2$. Continuing in this way, you get $H_n = \langle\alpha_1, \dots, \alpha_n\rangle$. Now what does property $N$ tell you about this sequence of subgroups and hence $G$?

Pick any element $g_1\in G$. Define $G_1 := \langle g_1\rangle$. If $G_1 = G$, we are done, and $G$ is finitely generated by just $g_1$. Otherwise, pick $g_2\in G - G_1$, and define $G_2 := \langle g_1,g_2\rangle$. If $G_2 = G$ then we are done, and $G$ is generated by $g_1,g_2$. Otherwise, continue doing this. If $G$ has property "$N$", then this process must stop after a finite number of steps, say it stops after step $k$, then we find that $G = G_k := \langle g_1,g_2,\ldots,g_k\rangle$, and $G$ is generated by $k$ elements.