Let $G$ be a finite group and $H$ be a subgroup of $G$. Suppose that for any prime power order element $x$ of $G$, there exists some element $g$ in $G$ such that $x^g$ is contained in $H$. Does it follow that $H=G$?

$\begingroup$ This question does not appear to be about research level mathematics within the scope defined in the help center. If this appearance is in error, you are encouraged to revise the question. meta.mathoverflow.net/questions/70/howtoaskpage $\endgroup$ – Theo JohnsonFreyd Nov 30 '13 at 6:43

6$\begingroup$ I think this is equivalent to asking whether, if we have a permutation group $G$ acting faithfully and transitively on a finite set $\Omega$, is it true that there is an element of prime power order which fixes no point? The answer to that question is "yes", but it is a difficult theorem, which requires the classification of finite simple groups. $\endgroup$ – Geoff Robinson Nov 30 '13 at 11:08

5$\begingroup$ So why was the question closed? $\endgroup$ – Derek Holt Nov 30 '13 at 11:27

2$\begingroup$ meta.mathoverflow.net/a/1220/2926 $\endgroup$ – Todd Trimble♦ Nov 30 '13 at 15:38

3$\begingroup$ @PeterMueller Your comment is substantive enough that it would be quite reasonable as an answer, IMO. $\endgroup$ – Todd Trimble♦ Nov 30 '13 at 17:00
As Geoff Robinson pointed out in his comment, the answer is yes by a deep and difficult theorem in finite groups. It was proven in Fein; Kantor; Schacher: Relative Brauer groups. II., J. Reine Angew. Math. 328 (1981), 39–57. As far as I know, no simpler proof has been found yet (maybe Theo JohnsonFreyd has a better one ...?).