# Why are Schur multipliers of finite simple groups so small?

Given a finite simple group $G$, we can consider the quasisimple extensions $\tilde G$ of $G$, that is to say central extensions which remain perfect. Some basic group cohomology (based on the standard trick of averaging a cocycle to try to make it into a coboundary) shows that up to isomorphism, there are only finitely many such quasisimple extensions, and they are all quotients of a maximal quasisimple extension, which is known as the universal cover of $G$, and is an extension of $G$ by a finite abelian group known as the Schur multiplier $H^2(G,{\bf C}^\times)$ of $G$ (or maybe it would be slightly more accurate to say that it is the Pontryagian dual of the Schur multiplier, although up to isomorphism the two groups coincide).

On going through the list of finite simple groups it is striking to me how small the Schur multipliers are for all of them; with the exception of the projective special linear groups $A_{n-1}(q)=PSL_n({\bf F}_q)$ and the projective special unitary groups ${}^2 A_{n-1}(q^2) = PSU_n({\bf F}_q)$, all other finite simple groups have Schur multiplier of order no larger than 12, and even the projective special linear and special unitary groups of rank $n-1$ do not have Schur multiplier of size larger than $n$ (other than a finite number of small exceptional cases, but even there the largest Schur multiplier size is 48). In particular, in all cases the Schur multiplier is much smaller than the order of the group itself (indeed it is always of order $O(\sqrt{\frac{\log|G|}{\log\log|G|}})$). For comparison, the standard proof of the finiteness of the Schur multiplier (based on showing that every $C^\times$-valued cocycle on $G$ is cohomologous to $|G|^{th}$ roots of unity) only gives the terrible upper bound of $|G|^{|G|}$ for the order of the multiplier.

In the case of finite simple groups of Lie type, one can think of the Schur multiplier as analogous to the notion of a fundamental group of a simple Lie group, which is similarly small (being the quotient of the weight lattice by the root lattice, it is no larger than $4$ in all cases except for the projective special linear group $PSL_n$, where it is of order $n$ at most). But this doesn't explain why the Schur multipliers for the alternating and sporadic groups are also so small. Intuitively, this is asserting that it is very difficult to make a non-trivial central extension of a finite simple group. Is there any known explanation (either heuristic, rigorous, or semi-rigorous) that helps explain why Schur multipliers of finite simple groups are small? For instance, are there results limiting the size of various group cohomology objects that would support (or at least be very consistent with) the smallness of Schur multipliers?

Ideally I would like an explanation that does not presuppose the classification of finite simple groups.

• I don't know if some intution may be derived from the following (I'm still struggling to try to understand the Schur multiplier), but there's the following: in "The second homology group of a group; relations among commutators" (Proc. Amer. Math. Soc. 3, (1952). 588–595) C. Miller shows that the second homology/Schur multiplier of $G$ can be interpreted as the group of all relations among formal commutators of elements of $G$, modulo those relations that hold "universally" (i.e., in the free group). I would expect few "nice" relations among commutators in simple groups beyond obvious ones. May 17, 2013 at 18:31
• There may be some insight gained from the transfer map from the homology of the Sylow subgroups. Maybe there's some general structure theory for Sylow subgroups of simple groups that shows that they have small Schur multiplier at each prime. May 17, 2013 at 19:02
• I remember asking Michael Aschbacher a similar question (I think it was actually why are the outer automorphism groups of all finite simple groups solvable) many years ago, and he opined that questions like that were pointless. Many common properties of the finite simple groups are just consequences of the classification. May 17, 2013 at 20:52
• @Terry Tao: Yes indeed; to give an explicit and unusual example with simple groups, the alternating group $A_{6}$ has an Abelian Sylow $3$-subgroup, but has a perfect triple cover. Less exotic are th example ${\rm PSL}(2,q)$ where $q \equiv \pm 3$ (mod 8) and $q>3.$ These have Abelian Sylow $2$-subgroups of order $4$, yet each has a perfect double cover ${\rm SL}(2,q).$ I think that Agol's suggested explanation is slightly off the mark: the Schur multipliers of Sylow $p$-subgroups of simple groups can get big, so I don't think you can explain the phenomenon locally (ie one prime at a time). May 24, 2013 at 13:56
• (1) An alternate measure of the size of an abelian group is rank; for Schur multipliers of finite simple groups, it is uniformly bounded by 2. Related, $Out(G)\to Aut(H_2(G))$, so a large rank Schur multiplier is a good candidate for a counterexample to the Schreier conjecture (only one simple group has non-solvable $Aut(H_2)$). So it may be hard to disentangle your question from the Schreier conjecture. (2) Perfect non-simple groups can have large Schur multipliers, certainly $|G|^c$ by just taking products of simples, but I think super-polynomial is possible. May 25, 2013 at 22:55

The Schur multiplier $H^2(G;{\mathbb C}^\times) \cong H^3(G;{\mathbb Z})$ of a finite group is a product of its $p$-primary parts

$$H^3(G;{\mathbb Z}) = \oplus_{ p | |G|} H^3(G;{\mathbb Z}_{(p)})$$

as is seen using the transfer. The $p$-primary part $H^3(G;{\mathbb Z}_{(p)})$ depends only of the $p$-local structure in $G$ i.e., the Sylow $p$-subgroup $S$ and information about how the subgroups of $S$ become conjugate or "fused" in $G$. (This data is also called the $p$-fusion system of $G$.)

More precisely, the Cartan-Eilenberg stable elements formula says that

$$H^3(G;{\mathbb Z}_{(p)}) = \{ x \in H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)} |res^S_V(x) \in H^3(V;{\mathbb Z}_{(p)})^{N_G(V)/C_G(V)}, V < S\}$$

One in fact only needs to check restriction to certain V above. E.g., if S is abelian the formula can be simplified to $H^3(G;{\mathbb Z}_{(p)}) = H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)}$ by an old theorem of Swan. (The superscript means taking invariants.) See e.g. section 10 of my paper linked HERE for some references.

Note that the fact that one only need primes p where G has non-cyclic Sylow $p$-subgroup follows from this formula, since $H^3(C_n;{\mathbb Z}_{(p)}) = 0$.

However, as Geoff Robinson remarks, the group $H^3(S;{\mathbb Z}_{(p)})$ can itself get fairly large as the $p$-rank of $S$ grows. However, $p$-fusion tends to save the day. The heuristics is:

Simple groups have, by virtue of simplicity, complicated $p$-fusion, which by the above formula tends to make $H^3(G;{\mathbb Z}_{(p)})$ small.

i.e., it becomes harder and harder to become invariant (or "stable") in the stable elements formula the more $p$-fusion there is. E.g., consider $M_{22} < M_{23}$ of index 23: $M_{22}$ has Schur multiplier of order 12 (one of the large ones!). However, the additional 2- and 3-fusion in $M_{23}$ makes its Schur multiplier trivial. Likewise $A_6$ has Schur multiplier of order 6, as Geoff alluded to, but the extra 3-fusion in $S_6$ cuts it down to order 2.

OK, as Geoff and others remarked, it is probably going to be hard to get sharp estimates without the classification of finite simple groups. But $p$-fusion may give an idea why its not so crazy to expect that they are "fairly small" compared to what one would expect from just looking at $|G|$...

I would be very surprised if you receive a "conceptual" answer to this problem- though I would be delighted to be proved wrong. Regarding your last comment, there have been examples recently where computational evidence has indicated that human intuition about the size of cohomology groups was probably faulty, being based on limited evidence.

Regarding the comment about the bad general bound for the size of the Schur multiplier of a finite group, it can get quite big for $p$-groups, as you no doubt know. If my memory is correct, an elementary Abelian $p$-group of order $p^{n}$ has Schur multiplier of order $p^{n(n-1)/2}$, as is well-known.