Recently, I am try to solve a problem in character theory:
character extension about $Q_8$
In this problem we have that $G=G/G'\cap N\lesssim G/G'\times G/N=G/G'\times Q_8$. If $G=N\times Q_8$, then $\theta$ extends to $G$.
So I need the following result in finite groups:
Let $G$ be a finite group and $N\trianglelefteq G$. Assume that $G\lesssim G/G'\times Q_8, G/N=Q_8$ and $N\le Z(G)$, then $G=N\times Q_8$.
Is it ture or not?
Take $G=C_4\rtimes C_4=\langle a,b\mid a^4=b^4=1, bab^{-1}=a^{-1} \rangle$ and let $N=\langle a^2b^2\rangle$.
The derived subgroup of $G$ is $\langle a^2\rangle$ with quotient isomorphic to $G/G'\cong C_4\times C_2$. $N$ is cyclic of order $2$, central in $G$ with quotient $Q_8$. $G$ is not isomorphic to $C_2\times Q_8$ but it does embed in $C_4\times C_2\times Q_8$, in fact in $C_4\times Q_8$.