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Let $p$ be an odd prime number. Can a finite simple group have a conjugacy class with $2p$ elements?

Leandro Vendramin
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    Did you trawl the ATLAS for examples? – Noam D. Elkies Sep 16 '11 at 03:30
  • Is it known that this can't occur with the alternating groups? Empirically it looks like the sizes of the conjugacy classes of $A_n$ all have at least 3 prime factors once $n>8$. – Alon Amit Sep 16 '11 at 06:17
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    I can see why a conjugacy class $C$ with $p$ elements is impossible: $G$ acts on $C$ by conjugation, faithfully as it is simple, so $G\subset S_p$. Under this embedding, the point-stabilizers $G\cap S_{p-1}$ are then the normalizers of elements of $C$. But every element normalizers itself, so the sizes of these subgroups must be a multiple of $p$, contradiction. This argument does not immediately work for $2p$, but it does put restrictions on $G\subset S_{2p}$ (e.g. it should contain the full $p$-Sylow $C_p\times C_p$ of $S_{2p}$). Maybe it can help to prove that this is impossible? – Tim Dokchitser Sep 16 '11 at 09:24
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    Actually, more generally a conjugacy class with $p^n$ elements is impossible for $n>0$. That's because if $C$ is a conjugacy class and $\chi$ is an irreducible char, and if $(\chi(1),|C|)=1$, then any $g\in C$ is either in the centre of $\chi$ or $\chi(C)=0$. If $G$ is simple, then the centres of characters are trivial, so $\chi(C)=0$ for all $\chi$ with $p\nmid \chi(1)$. The column orthogonality equation between 1 and $g\in C$ now gives a contradiction. This is somewhere in Isaacs, but for me, it's easier to link to my notes: http://math.postech.ac.kr/~bartel/docs/reptheory.pdf, Thm 6.7. – Alex B. Sep 16 '11 at 09:50
  • Unfortunately, this argument also doesn't immediately generalise to $|C|=2p$. – Alex B. Sep 16 '11 at 09:50
  • I would guess that this can't happen, because such an element gives a centralizer in a finite simple group $G$ of index $2p$. That is, you have a subgroup of small index with nontrivial center, which I think is not possible, after doing some examples in GAP and looking at the ATLAS. – Steve D Sep 16 '11 at 19:20

1 Answers1

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I believe the answer is no. A group with a conjugacy class of degree $2p$ has a transitive permutation action of degree $2p$ on that class, and the stabilizer is the centralizer of an element.

If this action is imprimitive, then the blocks have size $p$ or 2. Wtih blocks of size $p$ there would be a subgroup of index 2, contradicting simplicity. With blocks of size $p$, the group would act faithfully on the blocks. Burnside proved that a permutation group of prime degree $p$ is solvable or 2-transitive. The finite 2-transitive permutation groups are now known (using the classification of finite simple groups). They are listed, for example, in

PETER J. CAMERON, FINITE PERMUTATION GROUPS AND FINITE SIMPLE GROUPS, BULL. LONDON MATH. SOC , 13 (1981),1-22

and it can be checked that none of the examples of prime degree have point stabilisers that have a subgroup of index 2 with nontrivial centre.

So suppose that the action on the conjugacy class is primitive. In the Cameron paper cited above, a result of Wielandt is mentioned that says that primitive permutation groups of degree $2p$ either have rank 3 or are 2-transitive. Using the classification of finite simple groups, it has been checked that the only rank 3 examples are $A_5$ and $S_5$ in degree 10, and $A_5$ does not have a class of degree $2p$. For 2-transitive groups, we can again check the list and observe that none of the groups has point stabilizer with nontrivial centre.

Derek Holt
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