8
$\begingroup$

What is the Schur multiplier of the $n$-dimensional affine linear group $\mathrm{AGL}(n,q)$ over the Galois field with $q$ elements?

I am particularly interested in the simple case $n=1$. Computation using GAP shows that the Schur multiplier of $\mathrm{AGL}(1,q)$ is trivial for any prime power $q$ up to 19, except for $q=4$, in which case the multiplier has order 2. Is the Schur multiplier of $\mathrm{AGL}(1,q)$ always trivial except for $q=4$? Is the Schur multiplier of $\mathrm{AGL}(n,q)$ also trivial except for a few exceptions?

$\endgroup$
1
  • 2
    $\begingroup$ One explanation for the (so far, at least) exceptional ${\rm AGL}(1,4)$ multiplier is that this group is isomorphic with $A_4 \cong {\rm PSL}(2,3)$, which has a nontrivial double cover $2.A_4 \cong {\rm SL}(2,3)$. $\endgroup$ Jan 1, 2015 at 4:15

1 Answer 1

8
$\begingroup$

Let me concentrate on the ${\rm AGL}(1,q)$ case for now, where $q=p^k$ is a prime power. This group is a split extension $N:C$, where $N$ and $C$ are the additive and multiplicative groups of the field of order $q$, and the action of $C$ on $N$ is by multiplication in the field. So $N$ is elementary abelian and $C$ is cyclic.

Since cyclic groups have trivial multiplier, it follows from standard results that the $r$-part of the multiplier is trivial for all primes $r$ except possibly $r=p$ when $k>1$.

The Schur multiplier $M$ of $N$ is elementary abelian of order $p^{k(k-1)/2}$, and there is an induced action of $C$ on $M$. Again by standard theory, the $p$-part of the multiplier of $G$ is equal to the subgroup of $M$ that is fixed pointwise by $C$. (Since $|C|$ and $|M|$ are coprime, the action is completely reducible.)

If we regard $M$ and $N$ as ${\mathbb F}_pC$-modules, then $M$ is equal to the exterior square of $N$. So we need to determine whether $M$ has any trivial constituents as ${\mathbb F}_pC$-module.

If we extend to a splitting field for $N$ in characteristic $p$ (which we could take to be ${\mathbb F}_q$), then $N$ is isomorphic to the direct sum $\oplus_{i=0}^{p-1} N^{\sigma^i}$, where $N$ is now regarded as a module over ${\mathbb F}_q$, and $\sigma$ is the Frobenius field automorphism of ${\mathbb F}_q$ order $k$.

The exterior square of $N$ is a section of the tensor product $N \otimes N$. So if we can show that $N \otimes N$ has no trivial constituents, then we will have shown that the Schur multiplier of the group is trivial.

But, as a module over ${\mathbb F}_q$, $N \otimes N$ is the direct sum of $1$-dimensional submodules $N^{\sigma^i} \oplus N^{\sigma^j}$. The action of an element $g \in C$ on this submodule is by multiplication by $g^{\sigma^i+\sigma^j} = g^{p^i+p^j}$. So there will be a trivial constituent only if, for some $i$ and $j$ with $0 \le i,j < k$, we have $g^{p^i+p^j} = 1$ for all $g \in C$, which is equivalent to $p^k-1$ dividing $p^i+p^j$. It is easy to see that this happens only when $q=4$. (We are assuming that $k>1$.)

Continued on 5 January: Turning now to the case $G={\rm AGL}(n,q)$ for $n \ge 2$, we have $G = N:H$ with $H = {\rm GL}(n,q)$ and $N$ the natural module for $H$. Let $\hat{G}$ be a Schur covering group of $G$, so $\hat{G}/M \cong G$, where $M$ is the multiplier of $G$. Then $M$ is generated by the inverse images in $\hat{G}$ of $[H,H]$, $[G,N]$ and $[N,N]$, so let's consider those three subgroups in turn.

Firstly, (the inverse image of) $[H,H]$ is isomorphic to the Schur Multiplier of $H={\rm GL}(n,q)$. For $(n,q)=(2,2)$ and $(2,3)$, this is trivial. Otherwise $H' = {\rm SL}(n,q)$ is perfect, and $H/H'$ is cyclic, and it is not hard to see that the multiplier of $H$ is a quotient of that of $H'$. It is well-known that the multiplier of $H'$ is trivial except when $(n,q)$ = $(2,4)$, $(2,9)$, $(3,2)$, $(3,4)$ and $(4,2)$. It can be checked (for example by computer calculation) that the multiplier of $H$ has order $2$ when $(n,q)$ = $(2,4)$, $(3,2)$ or $(4,2)$, and is trivial otherwise.

Secondly, the inverse image of $[H,N]$ is isomorphic to ${\rm Ext}(N,T)$, where $T$ is the trivial module for $G$ over ${\mathbb F}_q$, which is isomorphic to $H^1(G,{\rm Hom}(N,T)) = H^1(G,\hat{N})$, where $\hat{N}$ is the dual module of $N$. Again these are in the literature for ${\rm SL}(n,q)$, and in the few cases where this is nonzero, it can be calculated for $H = {\rm GL}(n,q)$. It turns out that the only nonzero case is $(n,q) = (3,2)$, where the dimension is $1$.

Finally, the inverse image of $[N,N]$ is a quotient of the Schur Multiplier of $N$ and by a calculation similar to the case $n=1$, and using ${\rm AGL}(1,q^n) \le {\rm AGL}(n,q)$, we find that the only case when this is nontrivial is $(n,q)=(2,2)$ ($G \cong S_4$), when it has order $2$.

Summing up, the cases when the Schur multiplier of ${\rm AGL}(n,q)$ is nontrivial are $(n,q) = (2,2)$, $(2,4)$, $(3,2)$ and $(4,2)$, and it has order $4$ for ${\rm AGL}(3,2)$, and $2$ in the other cases. I have checked this with a computer calculation.

$\endgroup$
1
  • $\begingroup$ Thank you very much for the detailed explanation. Still the reasoning is a bit beyond my current understand. Is there a reference for the relevant results? $\endgroup$ Jan 8, 2015 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.