Why must we use Zorn's Lemma to prove every subgroup of $G$ is contained in a maximal subgroup of $G$?

Question

Why must Zorn's Lemma be used to prove every subgroup is contained in a maximal subgroup?

I understand the proof of it, but I feel like it's such an obvious statement that can be proved by common sense. If $H$ is a subgroup of $G$ and $H$ isn't maximal, there must exist subgroups in between them, so if we just look at any chain doesn't there have to be something "right under" $G$ in that chain (if we're looking at the subgroup lattice)?

I'm thinking that somehow Zorn's Lemma proves that there’s some "end" to the chain if the chain were infinitely long, but I still feel like common sense says there has to be an "end". I know there has to be something I am missing/incorrectly assuming, if anyone could help it would be much appreciated.

Answer 1

Zorn's Lemma alone does not suffice to deduce that statement for a general group as it is simply not true for an arbitrary group. A usual counterexample is the Prüfer $p$-group $\mathbb Z(p^\infty)$ which has as subgroups all cyclic groups of order $p^n$ which are increasingly ordered by inclusion. As there is a always $p^{n+1}$ after $p^n$ there is no maximal subgroup at all. This example is a special case of David A. Craven's comment if we only consider $p^n$-th roots of unity in $\mathbb C$.

However, two more remarks are to be made:

First, this proposition is indeed trivial (and requires no heavy machinery such as Zorn's Lemma) for finite groups. Here your approach works quite well as you have measure of "how far away from the top ($=$the whole group)" you are: the cardinality. In every sequence $H_1\subsetneq H_2\subsetneq\cdots\subsetneq G$, $|H_i|$ gets bigger and bigger but is bounded from above by $|G|$ and hence we can find a maximal subgroup in a reasonable way. However, this stops working once we consider arbitrary infinite groups; but see by second remark. As usual, infinities behave different from our intuition and we need tools like Zorn (or choice for that matter) to get some control back.

Second, Zorn's Lemma does (more or less) exactly what you say it does. Namely, it allows us to move up the whole chain of subgroups and find something on top. But there is an important observation: if we do the usual Zorn approach (i.e. consider the union of a chain to find upper bounds) we ran into the problem that we cannot guarentee that our union is different from the whole group, i.e. it might not be an upper bound at all!

To make things more formal. Consider a group $G$ with a subgroup $H$ and let

$$ \Sigma=\{M\,|\,H\leqslant M<G\}\,. $$

Now take a chain in $\Sigma$. The union over this chain will give us a subgroup containing $H$ but a priori we have no way of knowing if this union is still a strict subgroup, if this union is still in $\Sigma$! Without this control we cannot apply Zorn (and there is no way of fixing this as our counterexample(s) show).

Now take two different situations: "every (commutative) ring with $1$ has a maximal ideal" and "every finitely generated group has a maximal subgroup". Both of these are deducible using Zorn as we can use $1$ or the generators to control that our unions are not trivially the whole structure by choosing $\Sigma$ appropriately.

Answer 2

This isn't true, as noted by mrtaurho. But I have a simpler counterexample than the Prufer p-group.

For consider $\mathbb{Q}$ under the $+$ operation.

Let $G \subseteq \mathbb{Q}$ be a subgroup containing 1. Then I claim that $G$ cannot be maximal.

For in general, consider $\frac{a}{b}$ with $a$ and $b$ coprime. Then write $ca + db = 1$ for some integers $c, d$; then $\frac{1}{b} =\frac{a}{b} \cdot c + d$. So clearly, $\frac{a}{b} \in G$ if and only if $\frac{1}{b} \in G$.

Now consider $S =\{n \in \mathbb{Z}^+ \mid \frac{1}{n} \in G\}$.

Note that if $a$ is a factor of $b$ and $b \in S$, then $a \in S$. Also, note that if $a, b \in S$, then $lcm(a, b) \in S$. This is because $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$ will reduce in simplest form to a fraction with denominator $lcm(a, b)$.

Conversely, consider a set $S \subseteq \mathbb{Z}^+$ such that for all $n \in S$, if $b$ is a factor of $n$ then $b \in S$; and also such that for all $a, b \in S$, $lcm(a, b) \in S$; and also such that $1 \in S$. Call such a set an "ideal". Then define $G = \{\frac{w}{n} \mid n \in S, w \in \mathbb{Z}\}$. Then $G$ is a subgroup of $\mathbb{Q}$ containing 1.

Note that these two constructions are inverses which preserve the $\subseteq$ relation. So the partial order of ideals is isomorphic to the partial order of subgroups of $\mathbb{Q}$ containing 1.

In particular, suppose $G \subsetneq \mathbb{Q}$ and suppose that there is no group $G'$ strictly between $G$ and $\mathbb{Q}$.

Then $S \subsetneq \mathbb{Z}^+$, and there is no $S'$ strictly between $S$ and $\mathbb{Z}^+$.

But consider some $n \notin S$. Define $S' = \{k \in \mathbb{Z}^+ \mid k$ is a factor of $lcm(a, n)$, where $a \in S\}$. Note that $S'$ is an ideal. Also, suppose that $n^2 \in S'$. Then $n^2$ is a factor of $lcm(n, a)$ for some $a \in S$. Then, by analysing all prime factors of $n$, we see that $n^2$ is a factor of $a$. Then $n \in S$. Contradiction.

So we see that there is no maximal subgroup of $\mathbb{Q}$ which contains all of $\mathbb{Z}$.

Answer 3

Let me give a much more complicated solution, which compensates for that by having a very simple explanation.

Shelah proved that there is an uncountable group whose proper subgroups are all countable (see Does there exist any uncountable group , every proper subgroup of which is countable? for a discussion). Of course, this group cannot be abelian.

Now, here is a general fact:

Suppose that $G$ is a group, let $A$ be a countable subgroup of $G$ and $g\in G$, then $\langle A,g\rangle$ (the smallest subgroup generated by $A$ and $g$) is countable.

To see why, note that the elements of $\langle A,g\rangle$ are given by finite products of elements of $A$ with $g$ or its inverse and from an enumeration of $A$ we can produce an enumeration of $\langle A,g\rangle$.

Going back to Shelah's group, given any proper subgroup $A$, it is countable, so there is some $g\notin A$, and therefore $\langle A,g\rangle$ is a strictly larger subgroup, but being countable it is still a proper subgroup. Therefore $A$ was not maximal.

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In the choiceless situation, however, a counterexample can hit a lot closer to home. Namely, it is possible that there is a vector space over any fixed field whose proper subspaces are all finite dimensional, but the space itself is not. In that case you also have that there are no maximal subgroups.

This is in contrast to the case of vector spaces in $\sf ZFC$, where we can take any single vector, and look at a complement space to the one generated by that vector.