It suffices to show that every subrepresentation has a complement. Let $V$ be a representation of $G$ and $W \subseteq V$ a subrepresentation.

Step 1: Choose an arbitrary complement

As vector spaces, we can write $V = W \oplus W_0'$ for some subspace $W_0'$ (not necessarily invariant under $G$). Let $\pi_0: V \to W$ be the projection along $W_0'$: $\pi_0(w + w') = w \text{ for } w \in W, w' \in W_0'$

Note that $\pi_0^2 = \pi_0$ and $\text{im}(\pi_0) = W$, $\ker(\pi_0) = W_0'$.

Step 2: Average over the group

Define a new map by averaging: $\pi = \frac{1}{|G|} \sum_{g \in G} \rho(g) \circ \pi_0 \circ \rho(g^{-1})$

The division by $|G|$ is possible since $|G|$ is invertible in $\mathbb{F}$ (this is where we use $\text{char}(\mathbb{F}) \nmid |G|$).

Step 3: Verify $\pi$ is a projection

We need to show $\pi^2 = \pi$. Note that: $\pi^2 = \frac{1}{|G|^2} \sum_{g,h \in G} \rho(g) \circ \pi_0 \circ \rho(g^{-1}) \circ \rho(h) \circ \pi_0 \circ \rho(h^{-1})$

Substituting $k = g^{-1}h$: $= \frac{1}{|G|^2} \sum_{g,k \in G} \rho(g) \circ \pi_0 \circ \rho(k) \circ \pi_0 \circ \rho(k^{-1}) \circ \rho(g^{-1})$

Step 4: Simplify using $W$ is invariant

Since $W$ is a subrepresentation, $\rho(k)(W) \subseteq W$ for all $k$. For $w \in W$: $\pi_0(\rho(k)(w)) = \rho(k)(w)$ because $\rho(k)(w) \in W$ and $\pi_0$ projects onto $W$.

Therefore, $\pi_0 \circ \rho(k) \circ \pi_0 = \pi_0 \circ \rho(k)$ when restricted appropriately. This gives: $\pi^2 = \frac{1}{|G|} \sum_{g \in G} \rho(g) \circ \pi_0 \circ \rho(g^{-1}) = \pi$

Step 5: Verify $\pi$ is $G$-equivariant

For any $h \in G$: $\rho(h) \circ \pi = \frac{1}{|G|} \sum_{g \in G} \rho(h) \circ \rho(g) \circ \pi_0 \circ \rho(g^{-1})$ $= \frac{1}{|G|} \sum_{g \in G} \rho(hg) \circ \pi_0 \circ \rho((hg)^{-1}) \circ \rho(h)$

Since summing over all $g \in G$ is the same as summing over all $hg \in G$: $= \frac{1}{|G|} \sum_{g \in G} \rho(g) \circ \pi_0 \circ \rho(g^{-1}) \circ \rho(h) = \pi \circ \rho(h)$

Step 6: Construct the complement

Let $W' = \ker(\pi)$. Since $\pi$ is $G$-equivariant, $W'$ is a subrepresentation. We have: $V = \text{im}(\pi) \oplus \ker(\pi) = W \oplus W'$

This shows $W$ has a complementary subrepresentation.

Conclusion: By induction on dimension, every representation decomposes completely into irreducibles.