## Every nontrivial finitely generated group has maximal subgroups

This is an exercise using Zorn’s Lemma to prove that every nontrivial finitely generated group has maximal subgroups. Let $G$ be a finitely generated group, say $G = \langle g_1, g_2, \ldots g_n \rangle$, and let $\mathcal{S}$ be the set of all proper subgroups of $G$. $\mathcal{S}$ is partially ordered by $\subseteq$. Let $\mathcal{C} = \{ C_i \}_{i \in \mathbb{Z}}$ be an arbitrary chain in $\mathcal{S}$.

1. Prove that $H = \bigcup \mathcal{C}$ is a subgroup of $G$.
2. Prove that $H = \bigcup \mathcal{C}$ is a proper subgroup of $G$.
3. Apply Zorn’s Lemma to infer the existence of a maximal subgroup.

1. Let $x,y \in \mathcal{C}$. Now we have $x \in C_i$ and $y \in C_j$ for some $C_i$, $C_j$ in $\mathcal{C}$. Since $\mathcal{C}$ is a chain, either $C_i \subseteq C_j$ or $C_j \subseteq C_i$. Without loss of generality, suppose $C_j \subseteq C_i$. Then $xy^{-1} \in C_i$, so that $xy^{-1} \in \bigcup \mathcal{C}$. By the Subgroup Criterion, $\bigcup \mathcal{C}$ is a subgroup of $G$.
2. Suppose $\bigcup \mathcal{C} = G$. Then for each generator $g_i$, we have $g_i \in C_{k_i}$ for some integer $k_i$. Recall that every finite set of integers has a greatest element; say the greatest element of $\{ k_1, k_2, \ldots k_n \}$ is $m$. Then for all $i$, we have $g_i \in C_m$. But then $G \leq C_m$, so that $C_m$ is not proper. This is a contradiction, so $\bigcup \mathcal{C}$ is a proper subgroup of $G$.
3. By the previous two parts, every chain in $\mathcal{S}$ has an upper bound in $\mathcal{S}$. By Zorn’s Lemma, $\mathcal{S}$ has a maximal element which is by definition a maximal subgroup of $G$.