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 <G $ with the property that for every $H_i$ there exists $H_{i+1}$ such that $H_i < H_{i+1}$?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $H_1 < H_2 < \cdots < H_n < \cdots < G$.

Solution 4:

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 $G_i$.

Solution 5:

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.