Does there exist a finite group $G$ with normal subgroups $N_1$ and $N_2$ such that$N_1 \cong S_5$, $N_2\cong S_7$, $G/N_1\cong S_{42}$ and $G/N_2\cong S_{41}$?
Consider $N_1<N_2$. (Edit: Wrong since $N_2$ cannot have normal subgroup of order equal to that of $N_1$) $$S_{41}\cong G/N_2\cong \frac{G/N_1}{N_2/N_1}$$. I do not know structure of $S_7/S_5$.
The case $N_1\nless N_2$ remains.
On the other hand I tried to find $G$. One way to find example of a group with multiple properties is to consider multiple groups each with one property and consider their direct product. But $G=S_5\oplus S_7 \oplus H$, for some $H$, fails.
The fact that $41$ is prime may be useful.
The problem has many subgroups which makes it complicated. Please give a hint. Thanks!