# Does every infinite group have a maximal subgroup?

$$G$$ is an infinite group.

1. Is it necessary true that there exists a subgroup $$H$$ of $$G$$ and $$H$$ is maximal ?

2. Is it possible that there exists such series $$H_1 < H_2 < H_3 <\cdots with the property that for every $$H_i$$ there exists $$H_{i+1}$$ such that $$H_i < H_{i+1}$$?

• See Prüfer group. It doesn't have any maximal subgroups. Commented Jun 13, 2013 at 6:45
• Probably the most familiar example is $\mathbb{Q}$. Commented Jun 13, 2013 at 7:31

Rotman p. 324 problem 10.25:

The following conditions on an abelian group are equivalent:

• $$G$$ is divisible.

• Every nonzero quotient of $$G$$ is infinite; and

• $$G$$ has no maximal subgroups.

It is easy to see above points are equivalent. If you need the details, I can add them here.

• Well put, Babak! @MathsLover OP originally meant "original post" in a thread, which here would mean the posted question...but it has come to be used to refer to "the one who posted the original post/question" Commented Jun 14, 2013 at 0:54
• @BabakS. , thank you very much :) this is kind of you :) " i saw the comment yesterday but for some reason it's now deleted ! "
– FNH
Commented Jun 14, 2013 at 10:45
• @amWhy , nice this is nice shortcut ! English speakers have many many shortcuts with the words ! i remember that the first time i read word " gr8" in a comment on a website , it didn't give me any sense for few minutes !
– FNH
Commented Jun 14, 2013 at 10:47
• @MathsLover: I'll add it. Certainly. Commented Jun 15, 2013 at 5:17
• I have edited the statement from Rotman's book. The exercise is from the chapter on abelian groups, so he is implicitly assuming that the group is abelian. Otherwise the result does not hold (see this answer below). Commented Aug 21, 2020 at 9:15

As Prism states in the comments, the Prüfer group is an example of a group with no maximal subgroup. Define $\mathbb{Z}(p^{\infty})$ to be the set of all $p^n$-th roots of unity as $n$ ranges over the natural numbers. The operation is multiplication.

It can be shown that any subgroup of $\mathbb{Z}(p^{\infty})$ has the form $\mathbb{Z}/p^n\mathbb{Z}$, so the lattice of subgroups of $\mathbb{Z}(p^{\infty})$ is just the chain:

$$1\subset\mathbb{Z}/p\mathbb{Z}\subset\mathbb{Z}/p^2\mathbb{Z}\subset\mathbb{Z}/p^3\mathbb{Z}\subset\ldots\subset \mathbb{Z}(p^{\infty})$$

It follows that $\mathbb{Z}(p^{\infty})$ has no maximal subgroup. Since it is abelian, it has no maximal normal subgroup also.

• And $\mathbb Z(p^{\infty})$ has just this normal series. Commented Jun 13, 2013 at 7:08
• @Jared , i think that the definition of $Z(p^{\infty}$ is not obvious . also if we have proved that every subgroup of $Z(p^{\infty}$ have the form $Z/p^nZ$ , but how can we know that for every $n$ in positive integers , there is a subgroup which have the form of $Z/p^nZ$ ? is this also proved ? it's similar to : every subgroup of integers has the form $xZ$ for some $x$ and $xZ$ is always a subgroups of integers " $Z$ is integers here ".
– FNH
Commented Jun 14, 2013 at 20:24
• @MathsLover the subset of $p^n$-th roots of unity for some fixed $n$ is easily seen to be a group and straightforwardly isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$. Commented Dec 8, 2013 at 1:36

I like this example for its simplicity:

Let $A$ be any group with a proper subgroup $B$. Let $G = \prod_{i = 1}^{\infty}A$ and $H_n = \prod_{i = 1}^{n}A \times \prod_{i = n + 1}^{\infty}B$, then $H_1 < H_2 < \cdots < H_n < \cdots < G$.

• this is So nice example ! , thanx
– FNH
Commented Jun 13, 2013 at 9:03

Let for example $G$ be the dyadic rationals under addition, that is, all rationals of the form $\dfrac{a}{2^k}$, where $a$ ranges over the integers and $k$ ranges over the non-negative integers.

Then for any $i$, let $G_i$ be the set of integer multiples of $\dfrac{1}{2^i}$.

We can play the same game with $G$ the rationals under addition, with $G_i$ the set of integer multiples of $\dfrac{1}{i!}$.

Note that in both cases $G$ is the union of the $G_i$.

If the group $$G$$ is infinite and finitely generated then the answer is YES. Fix $$g\in G$$, then by Zorn's lemma there exists a (proper) maximal subgroup that contains $$g$$.

Here is a proof of the above. Let $$G=\langle x_1,\ldots,x_d\rangle$$ be finitely generated and take $$g\in G$$. Define the set $$\mathcal{M}=\{A\le G \mid g\in A,\ A\neq G\}$$, then for every chain $$A_1\le A_2 \le \ldots$$ in $$\mathcal{M}$$ you can take the union $$A=\bigcup_{n\in \mathbb{N}} A_n$$. Then $$A\le G$$, $$g\in A$$. If $$A=G$$, then $$x_1,\ldots,x_d \in A$$, but then we must have $$x_1,\ldots,x_d \in A_{l}$$ for some $$l\in \mathbb{N}$$, so $$A_{l}\notin \mathcal{M}$$, a contradiction. By Zorn's lemma there exists a maximal element in $$\mathcal{M}$$. But then $$M$$ is also maximal in $$G$$, as another subgroup containing $$M$$ also contains $$g$$ and $$M$$ was maximal w.r.t. containing $$g$$.

It is probably worth noticing that the equivalence with "divisibility" mentioned in other replies holds only for abelian groups.

EDIT: a bit more googling yielded that the answer in general is NO! This is a modification of Ol'shanskii monster. This is constructed in Theorem 35.2 of the book "Geometry of Defining Relations in Groups" (pg. 392)

Theorem 35.2 There are countable simple groups (that are moreover, Artinian of finite exponent) without maximal subgroups.

Hope this helps to shed some light on this.

• Is the down vote still valid after the edit? Commented Oct 19, 2017 at 21:02
• It looks invalid even without the EDIT section. Commented Oct 20, 2017 at 6:47
• @DerekHolt I am sorry, what looks invalid? I added a proof of my claim. Commented Oct 20, 2017 at 8:02
• The downvote! You asked whether the downvote was valid and I said that I did not think that it was. Commented Oct 20, 2017 at 14:08

In the same way, you have $$\mathfrak{S}_2 \subsetneq \mathfrak{S}_3 \subsetneq \cdots \subsetneq \mathfrak{S}_n \subsetneq \cdots \subsetneq \mathfrak{S}_{\infty}$$ where $\mathfrak{S}_{\infty}$ is the set of bijections $\mathbb{N}\to \mathbb{N}$ fixing all but finitely many numbers.

• Assuming you mean for $\mathfrak{S}_n$ to denote the set of bijections fixing all but $\{1, \dots, n\}$ sitting inside $\mathfrak{S}_\infty$, this is misleading. For instance, for every $n$, $\mathfrak{S}_n$ naturally sits inside the (proper) subgroup of bijections fixing all but finitely many numbers.
– bzc
Commented Jun 13, 2013 at 7:17
• Thank you, I edited my answer. In fact, the idea is to take the direct limit, here $\bigcup\limits_{n \geq 2} \mathfrak{S}_2$. Commented Jun 13, 2013 at 8:15
• Very nice observation so +1 Commented Jun 14, 2013 at 5:17