Let $G$ be a finite solvable group and $H$ be a normal subgroup of $G$. If all the involution of $H$ lie in a conjugacy class of $G$, then what can we say about the structure of $H$?
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$\begingroup$ That it is a group of even order? $\endgroup$– DonAntonioMar 4, 2016 at 11:51
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1$\begingroup$ This seems to be the same as asking about groups where the automorphism group acts transitively on the elements of order $2$ (by considering the holomorph of $H$). $\endgroup$– Tobias KildetoftMar 4, 2016 at 11:56
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$\begingroup$ Take $H$ to be any solvable group of even order, $p$ a prime not dividing $|H|$ and $G=H\times C_p$. $G$ and $H$ satisfy the conditions in question. We can only say that $H$ is solvable group of even order. $\endgroup$– p GroupsMar 4, 2016 at 13:16
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$\begingroup$ @pGroups We can't actually say that the order is even. $\endgroup$– Tobias KildetoftMar 4, 2016 at 13:21
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$\begingroup$ he is talking about involutions. (perhaps, you are considering groups without involutions too, is it?) $\endgroup$– p GroupsMar 4, 2016 at 13:22
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