$\{1\} =G_n\lhd G_{n-1}\lhd \cdots \lhd G_{1} \lhd G_{0} = G$ be the composition series of G and $K\lhd G$ Then after we eliminate equalities. Show that $\{1\} \lhd KG_n/K \lhd KG_{n-1}/K \lhd \cdots \lhd KG_1/K \lhd KG_0/K = G/K$ is a composition series of G/K
I want to prove it by contradiction. WLOG, $$(KG_0/K)/(KG_1/K)$$ is not simple. Then exists proper Q $$Q \lhd (KG_0/K)/(KG_1/K)$$ By 3rd isomorphism thm, we have $$\dot Q \lhd KG_0/KG_1$$ and, $$KG_1 \lhd \ddot Q \lhd KG_0 \ \ \ and\ \ \ \ddot Q/KG_1 \cong \dot Q$$ Then, I want to proper $$\ \dddot Q\ne G_1 or\ G_0\ s.t.\ G_1 \lhd \dddot Q \lhd G_0$$ Then we get contradiction. But I have no idea how to accomplish last step.
Hint: There is a canonical map $ \phi : G_0 \to KG_0 / KG_1 $, which is obviously surjective. Show that $ G_1 \subset \ker \phi $, so that the map descends to the quotient. What can you say about the homomorphic image of a simple group?