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The following question is from a past problem set in a course on group theory. For reference, the text used is by Derek Robinson, entitled "A Course in the Theory of Groups".

"Show that in a group $G$ whose order is some power of a prime $p$, $H$ is a maximal group if and only if $|G:H|=p$."

I am hoping for a solution to this. We were given the hint that in a group $G$ of order $p^k$ where $p$ is prime, if there is a proper subgroup $H$ of $G$, then $H$ is a proper normal subgroup of $N_G(H)$. This hint seems to be the key but I still don't see how.

Thank you in advance.

inkievoyd
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In nilpotent groups $G$, especially $p$-groups, "normalizers grow", that is if $H$ is a proper subgroup of $G$, then $H \subsetneq N_G(H)$. So if $H$ is maximal, then $G=N_G(H)$, that is, $H$ is normal. Hence $G/H$ cannot contain any subgroups, except the trivial ones. This can only happen when $|G/H|=p$. (You can use that if it is larger than $p$, $G/H$ has a non-trivial center ...).

Conversely, if $[G:H]=p$, then if $K$ is a subgroup with $H \subseteq K \subseteq G$, index$[G:K] \mid p$. So $K=H$ of $K=G$. Hence $H$ is maximal.

Nicky Hekster
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