Suppose $V$ is a representation of a finite group $G$ over a field $k$ of characteristic $0$, and suppose dim$V=3$ and $\wedge ^2V$ is reducible. Then $V$ is reducible.
I was trying to do it by contradiction. Suppose $V$ is irreducible, and as dim$\wedge^2V=3$, then either $\wedge^2V=V$, which is impossible as $V$ will be reducible, or $(\chi_V,\chi_{\wedge^2V})=0$. I do not know how to continue from here, any help would be great.Thank you.
I want to take advantage of the fact that $\wedge^3V$ is 1-dimensional.
We have a non-degenerate bilinear mapping $B:V\times \wedge^2V\to \wedge^3V$ defined by $$ B(v_1,v_2\wedge v_3)=v_1\wedge v_2\wedge v_3 $$ and extending linearly. Non-degeneracy follows from the facts that A) any non-zero vector of $V$ can be included in a basis $\{v_1=v,v_2,v_3\}$ and B) because $dim V=3<4$ any non-zero vector of $\wedge^2V$ is of the form $v_2\wedge v_3$ for some pair of linearly independent vectors $v_2,v_3\in V$. Furthermore, by the definition of the action of $G$ on the exterior powers we get that $$ B(gv,gv')=g B(v,v') $$ for all $g\in G, v\in V, v'\in \wedge^2V$.
If $M\subset \wedge^2V$ were a non-trivial subrepresentation, then I claim that $$ M^{\perp_B}:=\{v\in V\mid B(v,m)=0\ \text{for all $m\in M$}\} $$ would be a non-trivial subrepresentation of $V$. Non-degeneracy of $B$ implies that $\dim M^{\perp_B}=\dim \wedge^2V-\dim M=3-\dim M$. Also if $v\in M^{\perp_B}$ then for all $g\in G$ and all $m\in M$ we have $$ B(gv,m)=B(gv,gg^{-1}m)=gB(v,g^{-1}m)=g\cdot0=0. $$ This implies that $gv\in M^{\perp_B}$ proving the claim.