$ N_{1} $ , $ N_{2} $ are minimal normal subgroups of $ G $, $ G/N_{1}\cap N_{2} \cong G $?

Question

Let $ G $ is a finite group and $ N_{1} $ , $ N_{2} $ are minimal normal subgroups of $ G $ that $ N_{1} \neq N_{2} $. Suppose $ G/N_{1} $ and $ G/N_{2} $ are supersolvable. Then $ G $ is supersolvable ? $ G/N_{1}\cap N_{2} \cong G $ ?

Answer 1

Consider the homomorphism $\varphi$ from $G$ into $G/N_1\times G/N_2$, given by $\varphi\colon g\mapsto (N_1g,N_2g)$. Then $\ker\varphi=N_1\cap N_2=\langle 1\rangle$ (Arthur's proof above), so $G\cong G/(N_1\cap N_2)\cong {\rm Image}(\varphi)$. As subgroup of a direct product of supersolvable groups, ${\rm Image}(\varphi)$ is supersolvable, and thus $G$ is.