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If $P$ is the Sylow $2$-subgroup of a finite group $G$, $H <P$, and $x \in P$ so that no non-trivial element of $\langle x \rangle$ conjugates into $H$ (in $G$), and $|P|=|H||x|$, how can I show that $x$ acts on the cosets of $H$ in $G$ by an odd permutation?

EDIT: I don't understand (as in a comment) why no non-trivial element of $\langle x \rangle$ conjugating into $H$ implies that all cycles of $x$'s action on $H$'s cosets are of length $|x|$. Assuming this I know the answer to my original question. If someone could explain why this comment is true I would be very grateful.

Alexander Gruber
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user151882
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  • Consider the length of the cycles of $x$ in this permutation action, and the number of such cycles. – Geoff Robinson Oct 29 '14 at 20:26
  • Another hint. Try the case in which $H=1$ first; then generalise. – James Oct 29 '14 at 21:22
  • The assumption about no nontrivial element of $\langle x \rangle$ conjugating into $H$ is equivalent to saying that all cycles in the action of $x$ on the cosets of $H$ have length $|x|$. – Derek Holt Oct 29 '14 at 21:44
  • @DerekHolt I don't understand quite why this translates-- if you could explain that would be great. – user151882 Oct 29 '14 at 21:46
  • @GeoffRobinson I understand that $x$ induces a cycle of length $|x|$ of course, but don't quite understand why all cycles have that length. – user151882 Oct 29 '14 at 22:13

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