Exercise 2.4.17 (Every non trivial finitely generated group possesses maximal subgroups)

Proof. Let $G$ be a non trivial finitely generated group, say $G=⟨g_1,g_2,\dots ,g_n⟩$, and let $\mathcal{}S$ be the set of all proper subgroups of $G$. Then $\mathcal{}S$ is partially ordered by inclusion. Let $\mathcal{}C$ be a chain in $\mathcal{}S$.

Since $G$ is non trivial, $\{1_G\}$ is a proper subgroup of $G$, i.e., $\{1_G\}\in \mathcal{}S$, so $\mathcal{}S\ne \varnothing$.

Note that $\text{∅}$ is a chain (the empty chain), and any subgroup $K\in \mathcal{}S$ is an upper bound of $\text{∅}$ (the condition $\forall <!--\; FUNCTION\; APPLICATION\; -->L,L\in \varnothing \Rightarrow L\le K$ is vacuously true: since $L\in \varnothing$ is false, then $L\in \varnothing \Rightarrow L\le K$ is true).

Since the empty chain has an upper bound in $\mathcal{}S$, we may suppose in the following that $\mathcal{}C\ne \varnothing$.

(a)

We must suppose in this part that $\mathcal{}C\ne \varnothing$, because ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{H\in \varnothing }H=\varnothing$ is not a subgroup.

Consider the union $H=\underset{K\in \mathcal{}C}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}K$, where $\mathcal{}C\ne \varnothing$.

• Let $K_0\in \mathcal{}C$ be a subgroup of $G$: $K_0$ exists because $\mathcal{}C\ne \varnothing$. Then $1_G\in K_0$ and $K_0\subseteq \underset{K\in \mathcal{}C}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}K=H$, so $1_G\in H$ and $H\ne \varnothing$.
• Let $x,y$ be any elements in $H$. Then there are subgroups $K_1,K_2\in \mathcal{}S$ such that $x\in K_1$ and $y\in K_2$. Since $\mathcal{}C$ is a totally ordered set, $K_1\le K_2$ or $K_2\le K_1$. Suppose that $K_1\le K_2$ (the other case is similar). Then $x\in K_2$ and $y\in K_2$, therefore $x{y}^{-1}\in K_2$, and $K_2\subseteq H$, so $x{y}^{-1}\in H$.

In conclusion

$$H=\underset{K\in \mathcal{}C}{\bigcup }K\le G(\text{if }\mathcal{}C\ne \varnothing ).$$

(b)

Assume for the sake of contradiction that $H$ is not a proper subgroup, so that $H=G$. Then each $g_i$ must lie in $H=\underset{K\in \mathcal{}C}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}K$, therefore for each index $i\in [[1,n]]$, there exists some subgroup $K_i\in \mathcal{}C$ such that $g_i\in K_i$. Since $\mathcal{}C$ is totally ordered for inclusion, there is some index $i_0\in [[1,n]]$ such that $K_i\le K_{i_0}$ for all $i\in [[1,n]]$. Then $g_i\in K_{i_0}$ for all $i\in [[1,n]]$. This shows that $H=⟨g_1,g_2,\dots ,g_n⟩\le K_{i_0}$, where $K_{i_0}\le H$, so $H=K_{i_0}$. This is a contradiction, because by assumption, $H$ is not a proper subgroup of $G$, but $K_{i_0}$ is a proper subgroup.

This contradiction proves that $H=\underset{K\in \mathcal{}C}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}K$ is a proper subgroup of $G$.

(c)

First $S$ is a nonempty partially ordered set for inclusion. By the note in the preamble, the empty chain has an upper bound, and y part (a) and (b), every chain $\mathcal{}C\ne \varnothing$ has an upper bound, given by $H=\underset{K\in \mathcal{}C}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}K$. By the Zorn’s Lemma, $\mathcal{}S$ has a maximal element $M$. By definition of a maximal element, $M\in \mathcal{}S$, so $M$ is a proper subgroup, and if $M\subseteq K<G$, then $M=K$. Therefore $K$ is a maximal subgroup of $G$.

Every non trivial finitely generated group possesses maximal subgroups.