TITLE: Does every infinite group have a maximal subgroup?
QUESTION [13 upvotes]: $G$ is an infinite group.

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

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 $H_i < H_{i+1}$?

REPLY [5 votes]: 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.