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.
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