I know that $\rho: G\rightarrow GL(V)$ is a representation of the group $G$ iff $\rho$ is a homomorphism of groups. But, my professor use to do something shorter to prove that some function is or not a representation. That is the following:
Given a group $G$ with presentation (for example) $<a,b |~~ a^2=b^2=(ab)^2=1>$, to prove that some $\rho$ is a representation of $G$ we just need to verify that $\rho$ "satisfies the relations of the generators", i.e, that $\rho(a)^2=\rho(b)^2=\rho(ab)^2=1$(this one is the identical operator in $V$). And what I don't get is why this is valid.
Is it trivial that this implies that $\rho$ is a group homomorphism??
$G$ is the quotient of the free group $F$ on letters $a$ and $b$ by the smallest normal subgroup $N$ containing $a^2$, $b^2$ and $(ab)^2$.
Given any group $H$, and two elements $h_1$ and $h_2$, there is a unique homomorphism $\phi:F\to H$ with $\phi(a)=h_1$ and $\phi(b)=h_2$. In your example, $H=GL(V)$. If $h_1^2=h_2^2=(h_1h_2)^2=e$ then $a^2$, $b^2$ and $(ab)^2$ are in the kernel $K$ of $\phi$. This implies that $N\le K$ and so $\phi$ factors through $F/N=G$.