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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 $H^2(GL_n(F);V)=0$.

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.

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.

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

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

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 $H^p(GL_n(F)/(\mathbb{Z}/2\mathbb{Z});H^q(\mathbb{Z}/2\mathbb{Z};V))$. 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.

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 $H^2(GL_n(F);V)=0$.

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.

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

$1 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow GL_n(F) \rightarrow PGL_n(F) \rightarrow 1.$

Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$. The $E_2$-term is $H^p(PGL_n(F);H^q(\mathbb{Z}/2\mathbb{Z};V)$. 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.

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

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

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 $H^p(GL_n(F)/(\mathbb{Z}/2\mathbb{Z});H^q(\mathbb{Z}/2\mathbb{Z};V))$. 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.