modern treatment of the triviality of the multiplicator of the finite special linear group

Given your favorite version of the Heisenberg group one can prove the Stone-Von Neumann theorem. It is then not to hard to construct a family of representations on a central extension of $Sp\left(2n, F\right)$ by $\mathbb{C}^{\times}$, where for me $F$ is a local field.

In a past project I constructed such a representation in the case when $n=1$ and $F$ is finite, and I would like to deduce this special construction from the general construction outlined in my first paragraph. I believe this reduces to showing that when $F$ is finite $H^2 \left( SL_2 \left(F\right) , \mathbb{C}^{\times} \right)$ is trivial when $F$ has odd order not equal to 9.

That said, I've been bouncing back and forth between books and google trying to find a proof of the triviality of this Schur multiplier.Indeed I found one in Schur's 1904 paper

"Uber die Darstellung der endlichen Gruppen durch gebrochene lineare Substitutionen."

However I was hoping that in the more than 100 years since Schur published this there has been a less German and more modern treatment of the triviality of $H^2 \left(SL_2 \left(F\right), \mathbb{C}^{\times} \right)$. So I'm wondering, is there such a treatment? Perhaps, as an alternative to a reference, someone could provide a sketch of the proof.

Edit 1: In the Atlas of Groups one can find an algorithm to calculate the Schur multiplier of a finite group, given generators and relations. However, I'd hope that there is a less computational proof that better capitalizes on the specific nature of the group $SL_2 \left(F\right)$.

Edit 2:Geoff Robinson and Jack Schmidt have pointed out that the Schur multiplier is nontrivial in the case that $F$ has order 4 or 9. Hopefully, my revised question is answerable.

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Here is just a partial answer to indicate that one mostly knows the multiplier from basic group theory.

If G = SL( 2, q ) for some prime power q = pk, then the Sylow r-subgroups of G are either:

• Quaternion if r = 2 ≠ p,
• Cyclic if 2 ≠rp,
• Elementary abelian if r = p

In the first two cases, the Sylow r-subgroup already has trivial multiplier, so G has trivial r-multiplier, as in pages 107–108 and 117 of Schur (1907) or page 170 of Aschbacher's Finite Group Theory and pages 123, 151, 153 of Isaacs's Finite Group Theory.

In particular, the multiplier of G is always a p-group.

One can even calculate the multiplier of a nicer group, B, where the relations are easier to understand.

Caveat: I haven't finished the calculation, but perhaps here is one way to approach it.

Since the Sylow p-subgroup P is abelian, B = NG(P) controls the multiplier, that is, the multiplier of G is isomorphic to the multiplier of B. I think one could simply calculate the multiplier (start with the multiplier of P and the take into account the GF(q)× = B / P action, where it acts as squares; this is done on page 118–119 of Schur (1907)). Large k should have trivial multiplier because of the large action of GF(q)×, and small k should be trivial since the multiplier of P is already small. Apparently something goes wrong for medium sized k, at least when q = 4 or q = 9.

When q = 4, one gets B ≅ AGL(1,4) ≅ A4 which has SL(2,3) as a double cover. When q = 9, one actually gets an action of GF(q)× on an extra-special 3-group as a triple cover of B.

In general you would have to get:

$$\hat B = \langle a, b_i, z_{ij} : a^{q-1} = b_i^p = z_{ij}^p = 1, b^a = b^{(\zeta^2)}, [b_i,b_j]= z_{ij}, z_{ij} \in Z(\hat B) \rangle$$

and I guess the question is how a can act consistently on both b and z without forcing z = 1. This appears to be a matrix congruence over GF(q), and so is probably equivalent to Schur's method. Presumably it is also understandable as looking at alternating forms on GF(q)2, so related to your original Sp(2, F) problem.

At any rate here is a bibliography with some links.

• Schur, J. Über die Darstellung der endlichen Gruppen durch gebrochene lineare Substitutionen. J. für Math. 127, 20-50 (1904). JFM35.0155.01 URL:GDZPPN002165511

• Schur, J. Untersuchungen über die Darstellung der endlichen Gruppen durch gebrochene lineare Substitutionen. J. für Math. 132, 85-137 (1907). JFM38.0174.02 URL:GDZPPN00216633X

• Aschbacher, Michael. Finite group theory. Cambridge Studies in Advanced Mathematics, 10. Cambridge University Press, Cambridge, 1986. x+274 pp. ISBN: 0-521-30341-9 MR895134 Google books

• Isaacs, I. Martin Finite group theory. Graduate Studies in Mathematics, 92. American Mathematical Society, Providence, RI, 2008. xii+350 pp. MR2426855 Google books: ** p. 118 lemma, ** p. 123 quaternions have no multiplier, ** p. 151 def. of multiplier, ** p. 153 triviality of more multipliers.

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There are actually only 6 exceptions to when $SL(n,q)$ doesn't have trivial Schur multiplier. The hardest to compute (by hand, anyway) is probably $SL(3,4)$. In Steinberg's Lectures on Chevalley Groups (which you can download at www.math.wisc.edu/~ram/YaleNotes.pdf) he proves the case $q\neq 2$, $3$, $4$, or $9$ [pgs. 85ff in the PDF]. In chapter 7 of Karpilovsky's The Schur Multiplier, he proves the case $n>4$. So there are only a handful of possible exceptions. – user641 Aug 14 '11 at 9:14

The trouble is that there can be small exceptions to the general rule: for example, $A_{6} \cong {\rm PSL}(2,9)$, and $A_{6}$ has a triple cover. Hence the Schur multiplier of $A_{6}$ has order $6$, and ${\rm SL}(2,9)$ has a triple cover.

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SL(2,4) has SL(2,5) as a double cover too. I think SL(2,F) otherwise has a trivial Schur multiplier, but three other SL(n,F) are exceptional too. – Jack Schmidt Aug 12 '11 at 17:54
Oh sorry, I should probably have said I'm willing to rule out the cases $SL \left( 2,4 \right)$ and $SL \left(2,9\right)$. – jdmorgan Aug 12 '11 at 17:58