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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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. |
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