# A subgroup intersects conjugacy class of every prime power order element

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 $H=G$?

• 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/how-to-ask-page Nov 30 '13 at 6:43
• 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. Nov 30 '13 at 11:08
• So why was the question closed? Nov 30 '13 at 11:27
• meta.mathoverflow.net/a/1220/2926 Nov 30 '13 at 15:38
• @PeterMueller Your comment is substantive enough that it would be quite reasonable as an answer, IMO. Nov 30 '13 at 17:00