Proving that an $FG$-homomorphism is surjective

Question

Assume that $V$ is an $FG$-module.Prove that the subset $$V_0 = \{v \in V : vg = v \space \forall \space g \in G \}$$ is an $FG$- submodule of $V$. Also show that the function $$\phi: v \to \sum_{g \in G} vg \space \space (v \in V)$$ is an $FG$-homomorphism from $V$ to $V_0$. Is it necessarily surjective ?

I was able to prove that $V_0$ is an $FG$-submodule, and I was also able to prove that $\phi$ is an $FG$- homomorphism since if $h \in G$ we do have that $$\phi(vh) = \sum_{g \in G} vhg = \sum_{g \in G} vg = \sum_{g \in G} vgh$$ and so we have that $\phi(vh) = \phi(v) = \phi(v)h$.

But now I am stuck on how to prove whether $\phi$ is surjective or not ? I saw in the solution it was written that if $v \in V_0$ then $\phi(v \space / \space|G|) = v$ hence $\phi$ is surjective. But I don't understand that how is that possible or what it really means (Can you give a simple example ?)

Answer 1

Notice that if $v\in V_0$, we have $$\phi(v)=\sum_{g}{vg}=\sum_{g}{v}=|G|v$$ This is because we are summing $v$ with itself $|G|$ times. If we divide by $|G|$ we get a vector that hits $v$, so the map is surjective. (This only works if the characteristic of the field doesn't divide $|G|$; if it does then $\phi(v)=0$.)

Answer 2

For a simple example of when surjectivity doesn't hold, consider $F = V = G = \mathbb F_2$ and let $G$ act on $V$ trivially. Then it's easy to see that $V_0 = V$, and yet $\phi(0) = \phi(1) = 0$, hence $\phi$ is not surjective.