$ G/N $ is is a subgroup of a direct product of copies of the cyclic group $ C_{p-1} $ if $ p>2 $ Let $ G $ is soluble group and $ A $ be a unique minimal normal subgroup of $ G $. Then $ A $ is a elementary abelian group of a prime power. Let every chief factor of $ G/A $ has order $ 4 $ or a prime. $ \vert A \vert = p^{a} $, then $ \vert A \vert > 4 $ if $ p = 2 $ and $ \vert A \vert > p $ if $ p > 2 $. Suppose $ N = Fit(G) $. let $ Fit_{p}(G/A) = O_{p^{\prime}p}(G/A) = O_{p}(G/A) = N/A $. Hence $ Fit_{p}(G/A) = N/A $ is the intersection of the centralizers of the $ p $-chief factors of $ G/A $. Why $ G/N $ is is a subgroup of a direct product of copies of the cyclic group $ C_{p-1} $ if $ p>2 $ and $ G/N $ is a subgroup of a direct product of copies of $ S_{3} $ ? (every chief factor of $ G/N $ has $ 4 $ or a prime)

Let $C_i/A$, for $i=1,2,\ldots,k$, be the centralizers of the $p$-chief factors $K_i/L_i$ of $G/A$. Then $G/C_i \le {\rm Aut}(K_i/L_i)$, which is isomorphic to $C_{p-1}$ (i.e. the cyclic group of order $p-1$ - sorry for double use of $C$)  if $|K_i/L_i| = p$ or to $S_3$ if $|K_i/L_i|=4$.
Now $G/N = G/\cap_{i=1}^k C_i$ is isomorphic to a subgroup of $G/C_1 \times \cdots \times G/C_k$, which is itself isomorphic to a subgroup of ${\rm Aut}(K_1/L_1) \times \cdots \times {\rm Aut}(K_k/L_k)$.
