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Let $G$ be finite. Suppose that all maximal subgroups of $G$ are conjugate. Then $G$ is cyclic.

I was stuck, then I find one solution. In Jack’s answer, it was mentioned that “one conjugacy class of maximal subgroups in fact implies that there is only one maximal subgroup”.

However, in the beginning of the proof, Lagrange Theorem was applied to the conjugacy class $M$, which is not necessarily a subgroup. I’m really confused about this point. Any help is sincerely appreciated.

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In the answer you mentioned $M$ refers to an element of the conjugacy class rather than the conjugacy class itself. In that answer since each Sylow $p$-subgroup of $G$ is contained in a maximal subgroup (which is conjugate to $M$) while all Sylow $p$-subgroups are conjugate, it follows that $M$ contains a Sylow $p$-subgroup of $G$. Then the Lagrange theorem implies that $M$ has order divisible by every prime power dividing $|G|$. Thus $|G|$ divides $|M|$.

josephz
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