Let $P$ be a quaternion of order 8 and $Q$ a cyclic group of order 9 and $G=[p]Q$, a semidirect product ($P$ is normal in $G$).
Let $M$ be a maximal subgroup of $G$ such that $Q<M$. I want to find $|G : M|$=? in gap.
closed as not a real question by 5PM, Haskell Curry, Ittay Weiss, Davide Giraudo, Norbert Feb 2 '13 at 12:25
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There's really no need to use GAP.
There are two possibilities, according to the action of $Q$ on $P$.
If the action is trivial, then $M$ has index 2, and it is the product of $Q$ with one of the (cyclic) subgroups of $P$ of order $4$.
If the action is non-trivial, then $Q$ permutes cyclically the three subgroups of $P$ of order $4$. Then $M$ has index 4, and it is the product of $Q$ with the subgroup of order $2$ of $P$.