Let $G=P \ltimes Q$ be a finite group and $N$ be a normal subgroup of $P$ such that $P/N \cong D_8$. Let $a_1N,a_2N$ be two non-comuting involutions s.t. $P/N=\langle a_1N,a_2N \rangle$. If suppose that $B_j:= \langle a_j ,N \rangle$ and $[Q,B_j]=1$($j=1,2$), then i want to prove that $[P,Q]=1$.
MY WORK: In here, let $a \in P$ and $y \in Q$ be two arbitrary element. I need to show that $[a,y]=1$. By assumption, we have $[y,a_1]=[y,a_2]=1$. Now can i say $P=\langle a_1 , a_2 \rangle$ and conclude that $[P,Q]=1$?
If I understood correctly, I think you almost grabbed it at the end: I don't think you can say $\;P=\langle a_1,\,a_2\rangle\;$ , but you can definitely say $\;P=\langle a_1,\,a_2,\,N\rangle\;$, and now: since $\;[Q,B_j]=1\;$ , the commutators of $\;Q\;$ with $\;a_i\;$ and also with $\;N\;$ are trivial, and thus also are the commutators of $\;P\;$ with $\;Q\;$ , i.e.: $\;[P,Q]=1\;$