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Let $G$ be a finite nilpotent group and $M$ be a maximal subgroup of $G$. If $H$ is a proper non-trivial subgroup of $G$ such that $H\not\leq M$, then we can show that $H\cap M$ is a maximal subgroup of $H$. Because every maximal subgroup of $G$ is normal and using second isomorphism theorem implies that $|H:H\cap M|$ is prime.

Could we generalize this property for NON-nilpotent groups?

Any proof or counterexample will be appreciated! Thanks in advance.

According to spin's nice point let me add this assumption to the question:

"Let $H\cap M\neq 1$".

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  • $\begingroup$ There are many small counterexamples. I suggest you try looking for one. For example, you could look for examples where $H \cap M = 1$ but $H$ is not cyclic of prime order. $\endgroup$ – spin Nov 11 '14 at 16:28
  • $\begingroup$ To get $H \cap M \ne 1$, just take a direct product with any nontrivial group. $\endgroup$ – Derek Holt Nov 11 '14 at 21:26
  • $\begingroup$ Dear professor Holt please give me a little more details if it is possible. thank you so much! $\endgroup$ – shankfei Nov 12 '14 at 7:53

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