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Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.

Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)

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  • $\begingroup$ You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups. $\endgroup$ – Derek Holt Jul 15 '18 at 7:27
  • $\begingroup$ The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$) $\endgroup$ – Max Jul 15 '18 at 9:38
  • $\begingroup$ @Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $\mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $\mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $\mathbb S$ such that $\bigcup_{H\in S}H\notin \mathfrak L$, and hence $\bigcup_{H\in S}H=G$. Since $G$ is finitly generated, $G\in \mathbb S$. This is a contradiction as $G$ is not proper in $G$. $\endgroup$ – mesel Jul 15 '18 at 9:51
  • $\begingroup$ How do you deduce $G\in S$ ? $\endgroup$ – Max Jul 15 '18 at 10:21
  • $\begingroup$ @Max: $G$ is finitely generated, that is $\langle x_1,x_2,...,x_k\rangle=G$. As $\bigcup_{H\in \mathbb S}H=G$, set $x_i\in H_i$. Then $G=H_j$ for some $j$. Thus, $G\in \mathbb S$. $\endgroup$ – mesel Jul 15 '18 at 12:55
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No. For instance, let $G$ be a vector space over $\mathbb{F}_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.

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