Given the group $G=N\oplus M$, show $F(G)=F(N)\oplus F(M)$, where $F(G)$ denotes the Fitting group of $G$ (the product of all nilpotent normal subgroups).
I would say the inclusion $\supseteq$ is trivial, as $F(N)\oplus F(M)$ is itself a nilpotent normal subgroup of $G$, by characterization of the Fitting group it must be contained in $F(G)$, but the other inclusion is what worries me.
Best Answer
If $A\triangleleft G$ is a nilpotent normal subgroup of $G$, then $$\pi_N(A) = \{n\in N\mid \exists m\in M\text{ such that } (n,m)\in A\}$$ is a nilpotent normal subgroup of $N$ (being the image of a nilpotent normal subgroup under a surjective homomorphism), and likewise $\pi_M(A)$ is a nilpotent normal subgroup of $M$. Clearly, $A\leq \pi_N(A)\oplus \pi_M(A)\leq F(N)\oplus F(M)$. Thus, every nilpotent normal subgroup of $G$ is contained in $F(N)\oplus F(M)$.