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This never happens for finite fields $F \neq \mathbb{F}_2$. If a group $G$ acts on an abelian group $M$, then short exact sequences

$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$

are classified by elements of $H^2(G;M)$. It is thus enough to show that if $F \neq \mathbb{F}_2$ is a finite field and $V = F^n$, then $H^2(GL_n(F);V)=0$. In fact, we will show that $H^k(GL_n(F);V)=0$ for all $k$.

We have a short exact sequence

$1 \rightarrow F^{\times} \rightarrow GL_n(F) \rightarrow PGL_n(F) \rightarrow 1.$

Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$. The $E_2$-term is $H^p(PGL_n(F);H^q(F^{\times};V))$. The key fact here is that $H^q(F^{\times};V)=0$ for all $q$.

On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Define $N = \sum_{x \in F^{\times}} x \in \mathbb{Z}[F^{\times}]$ (of course, $N$ acts as $0$ on $V$, but forget that for the moment). We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{F^{\times}}$ and which satisfies $N(gv)=N(v)$ for all $g \in F^{\times}$ and all $v \in V$. Let $V_{F^{\times}}$ be the ring of coinvariants, ie the quotient of $V$ by the subspace spanned by $\langle g v-v\ |\ g \in F^{\times},\ v \in V\rangle$. We get an induced map $\overline{N}:V_{F^{\times}} \rightarrow V^{F^{\times}}$. The result then is that $H^0(F^{\times};V) = V^{F^{\times}}$, that $H^i(F^{\times};V) = ker\ \overline{N}$ for $i \geq 1$ odd, and that $H^i(F^{\times};V) = coker\ \overline{N}$ for $i \geq 1$ even. But clearly $V^{F^{\times}} = 0$, and since $F$ is not the field with $2$ elements we also have $V_{F^{\times}} = 0$. The result follows.

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This never happens for finite fields $q$ odd. F \neq \mathbb{F}_2$. If a group$G$acts on an abelian group$M$, then short exact sequences$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$are classified by elements of$H^2(G;M)$. It is thus enough to show that if$F$F \neq \mathbb{F}_2$ is a finite field of odd characteristic and if $V = F^n$, then $H^2(GL_n(F);V)=0$.

We have a short exact sequence

$1 \rightarrow \mathbb{Z}/2\mathbb{Z} F^{\times} \rightarrow GL_n(F) \rightarrow GL_n(F)/(\mathbb{Z}/2\mathbb{Z}PGL_n(F) \rightarrow 1,$

where $\mathbb{Z}/2\mathbb{Z}$ is generated by the matrix with $-1$'s on the diagonal.1.$Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in$V$. The$E_2$-term is$H^p(GL_n(F)/(\mathbb{Z}/2\mathbb{Z});H^q(\mathbb{Z}/2\mathbb{Z};V))$. H^p(PGL_n(F);H^q(F^{\times};V))$. The key fact here is that $H^q(\mathbb{Z}/2\mathbb{Z};V)=0$ H^q(F^{\times};V)=0$for all$q$. On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Let$t$be the generator of$\mathbb{Z}/2\mathbb{Z}$. Define$N = 1 + t$\sum_{x \in F^{\times}} x \in \mathbb{Z}[F^{\times}]$ (of course, $N$ acts as $0$ on $V$, but forget that for the moment). We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{\mathbb{Z}/2\mathbb{Z}}$ V^{F^{\times}}$and which satisfies$N(gv)=N(v)$for all$g \in \mathbb{Z}/2\mathbb{Z}$F^{\times}$ and all $v \in V$. Letting Let $V_{\mathbb{Z}/2\mathbb{Z}}$ V_{F^{\times}}$be the ring of coinvariants, we ie the quotient of$V$by the subspace spanned by$\langle g v-v\ |\ g \in F^{\times},\ v \in V\rangle$. We get an induced map$\overline{N}:V_{\mathbb{Z}/2\mathbb{Z}} \overline{N}:V_{F^{\times}} \rightarrow V^{\mathbb{Z}/2\mathbb{Z}}$V^{F^{\times}}$. The result then is that $H^0(\mathbb{Z}/2\mathbb{Z};V) H^0(F^{\times};V) = V^{\mathbb{Z}/2\mathbb{Z}}$V^{F^{\times}}$, that$H^i(\mathbb{Z}/2\mathbb{Z};V) H^i(F^{\times};V) = ker\ \overline{N}$for$i \geq 1$odd, and that$H^i(\mathbb{Z}/2\mathbb{Z};V) H^i(F^{\times};V) = coker\ \overline{N}$for$i \geq 1$even. But clearly$V^{F^{\times}} = 0$, and since$F$has odd characteristic, is not the field with$2$elements we also have that$V^{\mathbb{Z}/2\mathbb{Z}} = V_{\mathbb{Z}/2\mathbb{Z}} V_{F^{\times}} = 0$, and the . The result follows. 2 added 139 characters in body This never happens for$q$odd. If a group$G$acts on an abelian group$M$, then short exact sequences$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$are classified by elements of$H^2(G;M)$. It is thus enough to show that if$F$is a field of odd characteristic and if$V = F^n$, then$H^2(GL_n(F);V)=0$. We have a short exact sequence$1 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow GL_n(F) \rightarrow PGL_n(FGL_n(F)/(\mathbb{Z}/2\mathbb{Z}) \rightarrow 1.$1,$

where $\mathbb{Z}/2\mathbb{Z}$ is generated by the matrix with $-1$'s on the diagonal.

Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$. The $E_2$-term is $H^p(PGL_n(F);H^q(\mathbb{Z}/2\mathbb{Z};V)$. H^p(GL_n(F)/(\mathbb{Z}/2\mathbb{Z});H^q(\mathbb{Z}/2\mathbb{Z};V))$. The key fact here is that$H^q(\mathbb{Z}/2\mathbb{Z};V)=0$for all$q$. On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Let$t$be the generator of$\mathbb{Z}/2\mathbb{Z}$. Define$N = 1 + t$. We then get a map$N : V \rightarrow V$whose image lies in the ring of invariants$V^{\mathbb{Z}/2\mathbb{Z}}$and which satisfies$N(gv)=N(v)$for all$g \in \mathbb{Z}/2\mathbb{Z}$and all$v \in V$. Letting$V_{\mathbb{Z}/2\mathbb{Z}}$be the ring of coinvariants, we get an induced map$\overline{N}:V_{\mathbb{Z}/2\mathbb{Z}} \rightarrow V^{\mathbb{Z}/2\mathbb{Z}}$. The result then is that$H^0(\mathbb{Z}/2\mathbb{Z};V) = V^{\mathbb{Z}/2\mathbb{Z}}$, that$H^i(\mathbb{Z}/2\mathbb{Z};V) = ker\ \overline{N}$for$i \geq 1$odd, and that$H^i(\mathbb{Z}/2\mathbb{Z};V) = coker\ \overline{N}$for$i \geq 1$even. But since$F$has odd characteristic, we have that$V^{\mathbb{Z}/2\mathbb{Z}} = V_{\mathbb{Z}/2\mathbb{Z}} = 0\$, and the result follows.

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