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For $n \ge 8$ the Schur multiplier $H_2(BA_n, \mathbb{Z})$ (where $A_n$ denotes the alternating group) stabilizes to $\mathbb{Z}_2$, and hence there is a universal central extension $\widetilde{A}_n$ of $A_n$ by $\mathbb{Z}_2$.

Question 1: Do any interesting groups (e.g. central extensions of sporadic simple groups) occur as centralizers (of elements) in $\widetilde{A}_n$?

Question 2: What are their Schur multipliers?

Motivation (feel free to ignore): Write $X = B \widetilde{A}_n$. In trying to find a relatively concrete answer to this question, I was led to a map

$$H_3(X, \mathbb{Z}) \to \pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$$

which I believe is an isomorphism for sufficiently large $n$. In any case, this map, regarded as an element of $H^3(X, \mathbb{Z}_{24})$, transgresses to a class in

$$H^2(LX, \mathbb{Z}_{24})$$

where for a group $G$, $LBG$ is the (classifying space of the) adjoint quotient $G/G$, the groupoid whose objects are elements of $g$ and whose morphisms come from conjugation. In particular, this transgressed class restricts to a distinguished family of classes in

$$H^2(C(g), \mathbb{Z}_{24})$$

where $C(g)$ denotes the centralizer and $g \in \widetilde{A}_n$. This gives a distinguished family of central extensions of the groups $C(g)$ by $\mathbb{Z}_{24}$, and it seems interesting to ask what these central extensions are. In particular (feel extremely free to ignore) the Schur multipliers of the sporadic simple groups are all subgroups of $\mathbb{Z}_{24}$, and it would be nice if this construction in some sense explained that....

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Qiaochu Yuan
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  • The centralizer of a conjugacy class is a normal subgroup and hence is either $\tilde{A_n}$ or central. I guess it's not what you mean but the formulation is awkward. – YCor Feb 10 '15 at 07:03
  • @YCor: by the centralizer of a conjugacy class I mean the centralizer of an element in that conjugacy class (which is independent, up to isomorphism, of the choice of such an element, so its isomorphism class is a well-defined invariant of the conjugacy class). Sorry if that was unclear. – Qiaochu Yuan Feb 10 '15 at 07:05
  • IMHO the answer to Q1 is no, all these centralisers are boring, and look much the same as these in $A_n$, give or take a central extension of a semidirect product of a bunch of $A_k$ and $S_m$... – Dima Pasechnik Feb 10 '15 at 07:26
  • @Dima: yes, on second thought, that sounds about right, I think up to a $\mathbb{Z}_2$-central extension and then up to taking a subgroup of index $2$? – Qiaochu Yuan Feb 10 '15 at 07:43
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    You won't find anything at all interesting by looking at centralizers of elements of odd order. It's rather the opposite of what is asked, but double covers of $A_{n}$ do sometimes occur as involution centralizers in sporadic simple groups, eg the Lyons group Ly has an involution centralizer $\hat{A_{11}}.$ – Geoff Robinson Feb 10 '15 at 11:46
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    Concerning the motivation - in an answer to the question "third stable homotopy group of spheres via geometry?" I've mentioned work of Igusa from late 70ies which is related – მამუკა ჯიბლაძე Feb 10 '15 at 18:05

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The answer to Q1 is no, all these centralisers are boring, and look much the same as these in $A_n$ itself. Indeed, think what happens to them under the homomorphism squashing the central $\mathbb{Z}_2$.

Dima Pasechnik
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  • What do you mean? If you have $x$ and a lift $x'$, then clearly the centralizer of $x'$ is contained in the inverse image of the centralizer of $x$. But the other inclusion is unclear, could you elaborate? – YCor Feb 10 '15 at 08:25
  • But they have index at most $2$ in the inverse image of the centralizer of $x$. – Derek Holt Feb 10 '15 at 09:19
  • Ah I see, because $y\mapsto [x,y]$ is a homomorphism from the inverse image of the centralizer of $x$ to the center. – YCor Feb 10 '15 at 09:37