If $H\leq G$ and $H$ is a maximal subgroup of $G$ then $[G:H]$ is prime and vice versa

Question

Prove that

if $H$ is subgroup of $G$, $H$ is a maximal subgroup of $G$ iff $[G:H]$ is prime.

I was reading a proof and the proof used this theorem to prove the statement. It's the first time that I meet this theorem !

Can any one give me its proof ?

Answer 1

As pointed out by the other answers only one direction is true: If $H$ is not maximal then there is $K$ such that

$$H \subsetneqq K \subsetneqq G.$$

Then we get that $[G: H] = [G:K][K:H]$. Then condition of strict inclusion guarantees that each of these factors is not $1$. Thus $[G:H]$ is not prime.

Answer 2

This is true if you assume that $H$ is a normal subgroup.

If $[G:H]$ is prime, then you can prove that $H$ must be maximal by proceeding as in the other answer. If you know that $H$ is a normal maximal subgroup, then $[G:H]$ must be prime. This follows from the correspondence theorem: you can show that $G/H$ has no nontrivial proper subgroups and thus must be cyclic of prime order.

However, in general this statement is false. For example, in the alternating group $A_4$ subgroups of order $3$ are maximal and have index $4$, which is not prime.

For solvable groups we have the following result.

If $G$ is a finite solvable group and $M < G$ is a maximal subgroup, then $[G:M]$ is a power of a prime.

This is not true for groups that are not solvable, as the alternating group $A_5$ shows.

Answer 3

This is false for an arbitrary group $G$. If every maximal subgroup of a group $G$ has prime index, it follows that $G$ is solvable by a theorem of Hall. For example, $A_5$ has a maximal subgroup of index $6$.