If a finite group has exactly $2$ maximal subgroups, then it is $3$-cyclic

Question

Prove that if a finite group has exactly $2$ maximal subgroup, then it is $3$-cyclic.

I tried using the same method as here:

Finite groups with exactly one maximal subgroup

but it leads nowhere.

Answer 1

For finite groups, this below is maybe a partial answer, but provides you with some insight. You probably know that a group with only one maximal subgroup must be cyclic of prime power order. A group with exactly two maximal subgroups must also be cyclic and has precisely two different primes dividing its order:

Proposition If $G$ is a group with exactly two maximal subgroups, then it must be nilpotent.

Proof Let $M$ be one of the two maximal subgroups. We have to show that $M$ is normal in $G$. Since $M$ is maximal, we have either $G=N_G(M)$ (equivalent to $M \lhd G)$ or $M=N_G(M)$. In the latter case, and because $|G:N_G(M)|$ is exactly the number of conjugates of $M$, we have $|G:N_G(M)|=2=|G:M|$. And it is well-known that a subgroup of index $2$ is normal.

Note In this proposition one can replace "two" by $p$, where $p$ is the smallest prime dividing the order of $G$.

Proposition There exists no $p$-group which has exactly two maximal subgroups.

Proof Let $G$ be a $p$-group with exactly two maximal subgroups, and let $\Phi(G)$ be its Frattini subgroup. Then $G/\Phi(G)$ has exactly two maximal subgroups, namely the canonical images of the maximal subgroups of $G$. But $G/\Phi(G)$ is an elementary abelian group say of order $p^n$. If $n=1$ then $G/\Phi(G)$ is cyclic, implying $G$ is cyclic and a cyclic $p$-group has a unique maximal subgroup. Hence $n \geq 2$ and we can find an $N \lhd G$ with $G/N \cong C_p \times C_p$. But then $G/N$ has $p+1$ maximal subgroups and the preimages of those in $G$ yield a contradiction.

Corollary If $G$ is a group with exactly two maximal subgroups, then $G \cong C_{p^{a}q^{b}}$, with $p$ and $q$ two different primes.

For this $G$ has to be nilpotent, say $G=P_1 \times \cdots \times P_k$, the direct product of its different Sylow subgroups. It is not difficult to see that the number of maximal subgroups is the sum of the number of maximal subgroups of each of the factors. By the previous propositions and the fact that a $p$-group with only one maximal subgroup must be cyclic, the corollary now follows.

The last proposition also implies the following.

Proposition Let $p$ be an odd prime. There exists no $p$-group which has exactly three maximal subgroups.

Proof Let $M_1, M_2$ and $M_3$ be the maximal subgroups. These all have index $p$ and are normal. Put $N=M_1 \cap M_2$, and assume $N=1$. Then we have $G=M_1M_2$, so $|G|=p^2$, which group has $p+1$ maximal subgroups. So $p+1=3$, a contradiction. Observe that $N \subseteq M_3$, for if not, then $G/N$ has exactly two maximal subgroups, which is impossible by the previous proposition. So $N=M_1 \cap M_2 \cap M_3=\Phi(G)$. Hence $G/N \cong C_p \times \cdots \times C_p $ and this group has $\frac{p^n-1}{p-1}=3$ maximal subgroups (with $n \geq 2$). It follows that $p=2=n$ again a contradiction.

Note that $V_4$ and also $Q_8$, the quaternion group of order $8$, have exactly $3$ maximal subgroups. What remains is to classify all $2$-groups with exactly $3$ maximal subgroups.