most recent 30 from http://mathoverflow.net 2013-06-20T04:54:18Z http://mathoverflow.net/feeds/question/27033 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27033/determining-conjugacy-class-of-a-subgroup-from-intersection-with-conjugacy-classe Jamie Vicary 2010-06-04T09:57:28Z 2010-06-04T10:08:41Z

<p>Is a subgroup of a finite group uniquely determined, up to conjugation, by the subset of conjugacy classes of the larger group that it intersects?</p>

http://mathoverflow.net/questions/27033/determining-conjugacy-class-of-a-subgroup-from-intersection-with-conjugacy-classe/27034#27034 Robin Chapman 2010-06-04T10:08:41Z 2010-06-04T10:08:41Z

<p>Let $G$ be the group of affine linear maps over the Galois field $k=GF(16)$ of order $16$. The elements of $G$ are maps from $k$ to itself of the form $x\mapsto ax+b$ where $a\in k^*$ and $b\in G$. Those with $a=1$ form a normal elementary abelian subgroup~$H$. All nontrivial elements of $H$ are conjugate. Then $H$ contains lots of subgroups of order $4$, thirty-five in all, each consisting of the identity and three elements of this conjugacy class of involutions. But these are not all conjugate under $G$; it is clear that such a subgroup has at most fifteen conjugates.</p>