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?
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asked Jun 4, 2010 at 9:57
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$\begingroup$ This is certainly true when the subgroup intersects every conjugacy class, and when it intersects only one conjugacy class. See this recent related question: mathoverflow.net/questions/26979/… $\endgroup$– Jamie VicaryCommented Jun 4, 2010 at 9:59
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5$\begingroup$ The subgroups < (13)(24) > and < (13)(24), (12)(34) > in $S_4$ intersect the same conjugacy classes and aren't even isomorphic! $\endgroup$– Steve DCommented Jun 4, 2010 at 11:10
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$\begingroup$ Is it uniquely determined by the sizes of its intersections with each conjugacy class? $\endgroup$– VladimirCommented Nov 24, 2015 at 5:08
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1 Answer
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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.
answered Jun 4, 2010 at 10:08
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$\begingroup$ Thanks Robin! That's neat. Could you give me a presentation of that group so that I can play with this example myself? $\endgroup$ Commented Jun 4, 2010 at 10:42
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$\begingroup$ If you really want a presentation: generators $a$, $b$, $c$, $d$, $x$. Let $a^2=\cdots=d^2=x^{15}=1$, $a,\ldots,d$ commute, $xax^{-1}=b$, $xbx^{-1}=c$, $xcx^{-1}=d$, $xdx^{-1}=ab$. $\endgroup$ Commented Jun 4, 2010 at 10:53