Let $G_1,G_2,...$ be subgroups of a group $G$. I would like to show that if $G_i \subseteq G_{i+1},G_i \neq G_{i+1}$, then $\bigcup_{i=1}^{\infty} G_i$ is not a cyclic group.

This seems like an assume for the sake of contradiction proof, although I am having some difficulty finding it. Assume for the sake of contradiction $$\bigcup_{i=1}^{\infty} G_i = (x)$$ for some $x \in \bigcup_{i=1}^{\infty}.$ Then $\exists i \in \mathbb{N}. \forall j \geq i. x \in G_j.$ I'm wondering based on construction, if there is some $G_j = (x),$ and if I can declare some kind of contradiction based on this particular statement. It seems to me that $(x)$ would then be a proper subset of further $G_k$ for $k \geq j$, but I am not entirely sure if this is how I should go about finding the contradiction. Any suggestions on this problem?

  • $\begingroup$ More generally, and by almost the same proof, $\bigcup_{i=1}^\infty G_j$ is not a finitely generated group. $\endgroup$
    – Lee Mosher
    Jun 15, 2016 at 1:16

1 Answer 1


You are right. Under your assumption, $x\in G_j$ for some $j$. Then $(x)\subset G_j$. Also, $G_j\subset \cup_{i=1}^{\infty}G_i=(x)$, thus $G_j=(x)$. But since $G_i\subset G_{i+1}$, we have $(x)\subset G_{r}\subset \cup_{i=1}^{\infty}G_i=(x)$ for all $r>j$. Thus, $G_r=(x)$ for all $i>j$. However, according to the assumption $G_i\neq G_{i+1}$. $\Rightarrow\Leftarrow$.


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