Index of a maximal subgroup in a finite group

Solution 1:

If a group has the property that all its maximal subgroups have prime index, then it is solvable; this follows from a theorem of P. Hall.

In fact, Hall's theorem reaches that conclusion if the indices are prime or squares of primes. A later theorem of Huppert states that if all of them are prime, the group is supersolvable.

Solution 2:

No. The smallest counter-example I can think of is $A_5$, the alternating group on $5$ elements which has order $60$, which has a maximal subgroup of order $10$ and thus index $6$ in $A_5$ (more information can be found here).

However, this does hold for finite abelian groups. To see this, use the fundamental theorem of finite abelian groups to write out the abelian group $G$ as $\mathbb{Z}_{p_1^{e_1}}\oplus \mathbb{Z}_{p_2^{e_2}}\oplus \cdots \oplus \mathbb{Z}_{p_n^{e_n}}$. Each maximal subgroup of $G$ contains all of all but one group and a maximal subgroup of $\mathbb{Z}_{p_i^{e_i}}$ for some $i$. But the maximal subgroups of $\mathbb{Z}_{p_i^{e_i}}$ are of order $p_i^{e_i - 1}$, so the index of them in $\mathbb{Z}_{p_i^{e_i}}$ is $p_i$ and thus the index of the maximal subgroup of $G$ is $p_i$ as well.