Exercise A0.V.3.08

Proposition A

Every non-trivial finitely generated group has a maximal proper subgroup.

Proof

Let $G$ be a non-trivial finitely generated group. Let $\mathcal{C}$ be a chain of proper subgroups of $G,$ with $s_n \in \mathcal{C}$ and $s_0 \subseteq s_1 \subseteq s_2 \subseteq \ldots.$ Clearly $S = \bigcup \mathcal{C}$ is a subgroup of $G.$ Suppose $G \subseteq S.$ In this case, there must exist $n \in \mathbb{N}$ such that $\{g_0, g_1, \ldots, g_m\} \subseteq s_n,$ where $g_0,\ldots,g_m$ constitute a finite generating set of $G.$ In this case $s_n$ fails to be a proper subgroup of $G,$ contradicting our premises. Thus, $S$ is a proper subgroup of $G.$

Since every chain of proper subgroups of $G$ has an upper bound which is a proper subgroup of $G,$ Zorn’s lemma applies. Thus there exists a maximal proper subgroup of $G.$

$\blacksquare$

Proposition B

The group $(\mathbb{Q},+)$ has no maximal proper subgroup.

Proof

TODO

$\blacksquare$