Does every infinite group have a maximal subgroup?

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.

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.

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