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I have the dicyclic group $G$ of order 12 generated by $x,y$ satisfying $x^4 = y^3 = 1$ and $xyx^{-1} = y^2$, and am trying to determine whether the symmetric group $S_6$ contains a subgroup isomorphic to it.

So far I've tried looking for an appropriate set of 6 elements for $G$ to act on, and hoping that the permutation representation $ \phi: G \to S_6$ is injective, but haven't had any luck: $ \phi$ is not injective for the action of $G$ conjugating its set of 6 elements of order 4, nor is it injective for the action of $G$ translating its set of 6 cosets of a subgroup of order 2.

Is it even true that $S_6$ does contain such a subgroup isomorphic to $G$, and if so how would I construct the isomorphism?

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up vote 2 down vote accepted

If $y$ is of order $3$, it must have cycle type $(abc)$ or $(abc)(def)$. If $x$ is of order $4$, it must have cycle type $(abcd)$ or $(abcd)(ef)$. Also, you know how $x$ acts on $y$ by conjugation...

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I've tried working with that construction, and end up finding that in both cases, $x$ doesn't have the required cycle type. Am I doing something wrong? More specifically, if $y$ is a 3-cycle I end up finding $x$ has to fix one point, and if $y$ is a 3,3-cycle I end up with $x$ fixing two points or being a 6-cycle. – Jonas Nov 10 '12 at 5:03
@Jonas: I agree with everything you wrote; it sounds to me like it constitutes a proof that no such subgroup exists. – Micah Nov 10 '12 at 5:12
Ok, thanks for the help! Also, I meant to say "$x$ fixes two points and is a product of two transpositions" instead of just "$x$ fixes two points". – Jonas Nov 10 '12 at 5:16

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