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Prove or disprove: If $G$ is a finitely generated group which has a unique maximal subgroup, then $G$ is a finite cyclic group.

I am well aware that the result is true when $G$ is finite (see here). However, when $G$ is infinite, I am still uncertain whether the result is true or not. I tried to adapt the proof of the finite case as follows:

Let $M$ be the maximal subgroup of $G$. Take any $a\in G\setminus M$ and consider the subgroup $\langle a \rangle$. Under the hyphotesis that $G$ is finitely generated, if I could somehow prove that any proper subgroup is contained in a maximal subgroup, then $\langle a \rangle$ can not be proper and we are done. So we are left showing that any proper subgroup $H$ is contained in a maximal subgroup.

Suppose $G = \langle a_1,\dots, a_n\rangle$ and let $H\leq G$ be a proper subgroup. Set $S = \{M\leq G: M \mbox{ is proper and } H\leq M\}$. Any given chain in $S$ will have a maximal element in $S$, because $G$ is finitely generated. From Zorn's lemma, we see that $S$ has a maximal element. However, I still cannot guarantee that such maximal element will be maximal within $G$.

From this point onwards I dunno how to proceed. As of now, I believe the result is probably false, but I cannot come up with a good counter-example, since my knowledge of infinite groups is by no means deep.

Any help with a counter-example or how to correct the proof is well appreciated. Thank you

user2345678
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    Maximal elements of $S$ are maximal among the subgroups of $G$. If $K$ is maximal of $S$ and $K \leq L \leq G$, then, in particular $H\leq L$. So, $L\in S$ or $L=G$. By the maximality of $K$, then $K=L$ or $L=G$. – NDB Jul 10 '23 at 14:31

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Let $H < G$ be a proper subgroup. Choose $i$ with $1 \le i \le n$ such that $\langle H,a_1,\ldots,a_{i-1} \rangle \ne G$ and $\langle H,a_1,\ldots,a_i\rangle=G$.

By Zorn's Lemma there is a subgroup $M$ of $G$ that is maximal subject to $\langle H,a_1,\ldots,a_{i-1} \rangle \le M$ and $a_i \not\in M$. Then $M$ is maximal in $G$.

Derek Holt
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