If the following question is not good enough, please close it.


Please give a elementary proof or reference of the following result:


Let $G$ be a finite groups of order $120$, and $|Z(G)|=2$. Suppose $G/Z(G) \cong A\_5$ and $G'=G$, then $G \cong SL(2,5)$.


Of course, using the Schur index of $A\_5$, I can get the proof. But how to prove a group to be a classical group is always trouble for me. I know this result of may be well-known for a long time ago, but I can not find the proof in any textbook. (I study this because that I want to know more about groups of order $8pq$.



about the non-solvable group of order $120$
Since you know how to prove this already, it is difficult to know what you are looking for!


Here is a quick proof that there is a unique isomorphism class of groups $G$ with this property using the well-known fact that $A\_5$ has the presentation $\langle x,y \mid x^2=y^3=(xy)^5=1 \rangle$.


Let $t$ generate the centre of your group $G$, and let $X,Y$ be inverse images in $G$ of generators $x,y$ of $A\_5$. By replacing $Y$ by $Yt$ if necessary, we can assume that $Y^3=1$. By replacing $X$ by $Xt$ if necessary we can assume that $(XY)^5=1$. If $X^2=1$, then $X$ and $Y$ generate a subgroup isomorphic to $A\_5$, which must be a complement of $Z(G)$, contradicting $G'=G$. So $X^2=t$ and hence


$$G \cong \langle X,Y,t \mid t^2=1, X^2=t, y^3=(XY)^5=1, tY=Yt \rangle.$$


