# Simple groups and Schur multiplier

Let $G/K$ be a simple finite non-abelian group. We know that $G/K\cong G^{\prime}/(G^{\prime}\cap K)$. Now I would like to prove $G^{\prime}\cap K$ is a homomorphic image of the Schur multiplier of the simple group $G/K$.

• Since $G/K$ is a non-abelian simple group, so we have $G/K=(G/K)^{\prime}$. – Tina Jun 14 '14 at 9:42
• You need to assume that $K \le Z(G)$. If you assume that, then it's true for any group $G$: there is no need to assume that $G/K$ is simple or that $G$ is finite. – Derek Holt Jun 14 '14 at 10:40
• Thanks a million for your reply. – Tina Jun 14 '14 at 11:15

Just to add a bit more detail to my comment, let $G$ be and group and $K \le Z(G)$. Let $\phi:F \to G$ be a surjective mapping with $F$ free and let $R = \phi^{-1}(K)$, so $F/R \cong G/K$. Since $K \le Z(G)$, $[F,R] \le \ker (\phi)$, so $\phi$ induces $\bar{\phi}:F/[F,R] \to G$.
Now $\phi(F') = G'$ and, since $R = \phi^{-1}(K)$, $\phi(F' \cap R) = G' \cap K$, so $\bar{\phi}$ maps $(F' \cap R)/[F,R]$ onto $G' \cap K$, and hence $G' \cap K$ is a homomorphic image of $(F' \cap R)/[F,R]$, which is the Hopf formula for the Schur Multiplier of $F/R \cong G/K$.