Fitting group of a direct product

Question

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.

Answer 1

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 $F(N)\oplus F(M)$.

Answer 2

For future readers: The answer by Arturo Magidin inspired me to look at this in another way: One checks that (if I'm not mistaken) that $$0\rightarrow F(N)\rightarrow F(G)\rightarrow F(M)\rightarrow 0$$ with the canonical injection $\iota_N:F(N)\rightarrow F(G)$ and the projection $\pi_M:F(G)\rightarrow F(M)$ described by Arturo Magidin is a short exact sequence. Then $\pi_N$ is a retract for the injection $\iota_N$, so the sequence splits and the splitting lemma gives the desired result.