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?

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    $\begingroup$ 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 $2.A_4 \cong {\rm SL}(2,3)$. $\endgroup$ – Noam D. Elkies Jan 1 '15 at 4:15

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.

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  • $\begingroup$ 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? $\endgroup$ – Huangjun Zhu Jan 8 '15 at 17:51

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