Extensions of a projective special linear group Is it possible to classify groups that can be decomposed as a semidirect product $G=\mathrm{PSL}_2(q) \rtimes \langle t\rangle$, such that $t\in G$ has order $4$ and $Z(G)=\langle t^2\rangle$?
 A: These groups can be written as subgroups of $Aut(PSL_2(q)) \times C_4$. Note that $Aut(PSL_2(q)=P\Gamma L_2(q)$ and, writing $q=p^f$,  that 
$$
P\Gamma L_2(q)/ PSL_2(q) \cong  \begin{cases}
C_2 \times C_f, & \textrm{ if $q$ is odd};\\
C_f, & \textrm{ if $q$ is even}. 
\end{cases}
$$
These facts are enough to work out that your group $G$ will be generated by $PSL_2(q) \times \{1\}$ and an element $(\phi, g)$ where $\phi$ is an outer involutory automorphism of $PSL_2(q)$ and $g$ generates $C_4$. Now the possibilities for $G$ just depend on what the possibilities for $\phi$ are.


*

*If $q$ is an odd square, then you can choose $\phi$ to be either diagonal or a field automorphism; in the former case, the group $G$ has structure $2.PGL_2(q)$; in the latter the group $G$ is a subgroup of $2.P\Sigma L_2(q)$.

*If $q$ is an even square, then $\phi$ must be a field automorphism, and the group $G$ is a subgroup of $2.P\Sigma L_2(q)$.

*If $q$ is an odd non-square, then $\phi$ must be diagonal, and the group $G$ has structure $2.PGL_2(q)$.

*If $q$ is an even non-square, then no such group exists (since there is no choice of $\phi$ available).
Note that in the first case -- $q$ an odd square -- you might be wondering why $\phi$ cannot be a product of a field and a diagonal automorphism. The reason is that there is no such product of order $2$ -- the extension of $PSL_2(q)$ that you obtain by adjoining such an automorphism is non-split.
In this case, though, you can choose $\phi$ to have order $4$ in which case (as Derek Holt points out in his comment above), the group generated by $PSL_2(q)\times\{1\}$ and $(\phi, g)$ satisfies all of the properties that you require, except that $Z(G)$ is not equal to $\langle t^2\rangle$, although it is of order $2$.
