# Semi directProduct and Maximal subgroup in gap [closed]

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.

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## closed as not a real question by 5PM, Haskell Curry, Ittay Weiss, Davide Giraudo, NorbertFeb 2 '13 at 12:25

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Please try and format the question adequately, and more importantly, please complete it! –  Andreas Caranti Jan 31 '13 at 13:44
Hello Dr.andrea caranti My question is wrong ? –  M.Mazoo Feb 1 '13 at 20:02
@maghsoudmazoochian What are you asking? I understand $G=PQ$ and $M$ is a maximal subgroup of $G$. What do you want to know? –  Alexander Gruber Feb 1 '13 at 22:49
I do not fully understand the meaning of the question, but I have edited your post to use LaTeX as well as to fix your missing question statement. Here, square brackets on their own line are used to list urls, which is why your remark on what they meant did not appear. –  Muphrid Feb 2 '13 at 7:41
OK, may I safely assume $Q$ acts non-trivially on $Q$? (To define a semidirect product of $P$ by $Q$, you have to specify a group morphism $Q \to \operatorname{Aut}(P)$.) –  Andreas Caranti Feb 2 '13 at 7:46

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 $P$.