Let $G$ be a finite group such that $G$ has a normal subgroup $H$ and $H$ is isomorphic to the alternating group $A\_5$. Also we know that $G/H \cong A\_5$.


Can we say that $G \cong A\_5\times A\_5$?
Thanks for your helps



on the extensions of $ A_5$ by $A_5$
Well, $G$ must be isomorphic to the direct product $A\_{5} \times A\_{5}$ in any case, under your assumptions, since the outer automorphism group of $A\_{5}$ just has order $2$. Note that $H$ is a maximal normal subgroup of $G,$ and that $F(G) = 1$ under your assumptions. If $G \not \cong A\_{5} \times A\_{5},$ then $H = F^{\*}(G),$ and $G/H$ is isomorphic to a subgroup of ${\rm Out}(H)$ (it is always the case for any finite group $X$ that $X/F^{\*}(X)$ is isomorphic to a subgroup of ${\rm Out}(F^{\*}(X)).$


Here is a more direct proof without using prior knowledge of ${\rm Aut}(A\_{5})$ : we have $G = HN\_{G}(P)$ for a Sylow $5$-subgroup $P$ of $G.$ Hence $[N\_{G}(P):N\_{H}(P)] = [G:H] = 60.$ There is an element $x \in N\_{G}(P) \backslash N\_{H}(P)$ whose order is a power of $3.$ Now $x$ centralizes $P,$ since $[N\_{G}(P):C\_{G}(P)]$ divides $4$. Since $H$ has $6$ Sylow $5$-subgroups, and $x$ permutes the Sylow $5$-subgroups of $H$ by conjugation, and already fixes $P,$ we see that $x$ must normalize at least one more Sylow $5$-subgroup of $H,$ say $Q.$ As with $P,$ we see that $x$ must centralize $Q$ too. Then $x$ centralizes $\langle P,Q \rangle - H$ ( note that a subgroup of $H$ with more than one Sylow $5$-subgroup has at least 6, so has order at least $30,$ whereas $H$ has no subgroup of index $2$. Thus $HC\_{G}(H) >H,$ so that $G = H \times C\_{G}(H)$ as $G/H$ is simple.


