I want to solve the following problem from Dummit & Foote's Abstract Algebra:
This is exercise involving Zorn's Lemma (see Appendix I) to prove that every nontrivial finitely generated group possesses maximal subgroups. Let $G$ be a finitely generated group, say $G=\langle g_1,g_2,\dots,g_n \rangle$, and let $\mathcal{S}$ be the set of all proper subgroups $G$. Then $\mathcal{S}$ is partially ordered by inclusion. Let $\mathcal{C}$ be a chain in $\mathcal{S}$.
(a) Prove that the union, $H$, of all the subgroups in $\mathcal{C}$ is a subgroup of $G$.
(b) Prove that $H$ is a proper subgroup. [If not, each $g_i$ must lie in $H$ and so must lie in some element of the chain $\mathcal{C}$. Use the definition of a chain to arrive at a contradiction.]
(c) Use Zorn's Lemma to show that $\mathcal{S}$ has a maximal element (which is, by definition, a maximal subgroup).
Here is my attempt at a solution:
REMARK: The goal of this exercise is to prove that any chain has an upper bound in $\mathcal{S}$. It seems that the case of an empty chain needs to be handled separately: The empty chain $\mathcal{C}=\emptyset$, has any element of $\mathcal{S}$ as an upper bound. From now on, we assume all chains are nonempty.
(a) Let $\mathcal{C}$ be a nonempty chain in $\mathcal{S}$, and let $H=\cup_{K \in \mathcal{C}} K$. Since $\mathcal{C}$ is nonempty, it contains some subgroup $K_0$, which has the identity element. Thus $1 \in H$, so $H \neq \emptyset$. Suppose $x,y \in H$ then there are subgroups $K_1,K_2 < G$ in $\mathcal{C}$, such that $x \in K_1,y \in K_2$. Since $\mathcal{C}$ is a chain we must have either $K_1 \subseteq K_2$ or $K_2 \subseteq K_1$. Suppose WLOG that $K_1 \subseteq K_2$. Then both $x,y$ are elements of $K_2$, and since it's a subgroup we have $xy^{-1} \in K_2 \subseteq H$. Thus $H$ is a subgroup of $G$.
(b) Suppose, by way of contradiction that $H$ is not a proper subgroup. Then there exist subgroups $K_1,\dots,K_n \in \mathcal{C}$ such that $g_i \in K_i$ for $1 \leq i \leq n$. As before, it follows from the definition of the chain that one of the subgroups $K_1,K_2$ contains both of $g_1$ and $g_2$. Similarly one of the subgroups $K_1, \dots, K_n$ must contain all generators $g_1,\dots,g_n$ of $G$. Suppose WLOG that $K_n$ is that subgroup, we arrive at a contradiction since any element of $G$ may be constructed from the generators, so $K_n=G$ (This is a contradiction, because the elements in the chain are proper subgroups).
(c) Combining the previous two parts, we have shown that any (nonempty) chain in $\mathcal{S}$ has an upper bound in $\mathcal{S}$ (the empty case was taken care of in the remark). Thus Zorn's lemma furnishes the existence of a maximal element in $\mathcal{S}$. By definition this is a proper subgroup $M <G $, such that if $M \subseteq K$ for any other proper subgroup $K \in \mathcal{S}$, then $M=K$. Thus there cannot be any subgroup $K<G$ which is a proper supergroup of $M$. In other words, $M$ is a maximal subgroup of $G$.
Is my proof correct? Must the case of an empty chain be handled separately? Thank you!
