Given that $G$ is a finite group, prove that $F(G/Z(G))$ = $F(G) / Z(G)$. I have seen the equation in the title while reading about the Fitting subgroup, but I can't work out why those two quantities are equal.

Could anyone please explain the proof?

Thank you very much.
I know that $F(G)$ is the product of all $O_p(G)$ for all primes $p$.

Is it true that $O_p(G)/Z(G)$ = $O_p(G/Z)$ for each prime $p$?


The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H \leq F(G)$, which proves the other direction we wanted.
