Semidirect Product $(A_{5} \times A_{5}) \rtimes Z_{2}$ Consider the group $G=(A_{5} \times A_{5}) \rtimes Z_{2}$, where $A_{5} \times A_{5}$ is the normal subgroup of $G$. $Z_{2}$ acts by swapping the two copies of $A_{5}$.
I checked with gap that $A_{5} \times A_{5}$ is the unique minimal normal subgroup of $G$. Is it possible to prove that without calculation?
$A_{5}$ is the alternating group of degree $5$ and $Z_{2}$ is the cyclic subgroup of order $2$.
 A: I assume $Z_2$ acts by swapping the two copies of $A_5$. Let $K = A_5 \times A_5$. Write $K_1$ and $K_2$ for the two copies of $A_5$, that are the components of the direct product.
Consider first the normal subgroups of $K$. We want to show that they are $1, K_1, K_2, K$. Since $A_5$ is simple, $1 < K_1 < K$ is a chief series of $G$, a (nontrivial, proper) normal subgroup $M$ of $K$ must be isomorphic to $A_5$. If $M \ne K_1$, then $M$ and $K_1$ intersect trivially (because they are both simple and normal, and thus minimal normal in $K$), and thus they commute. Since $C_K(K_1) = K_2$, we have $M = K_2$.
Now since $Z_2$ swaps $K_1$ and $K_2$, the only nontrivial normal subgroup of $G$ contained in $K$ must be $K$ itself, so $K$ is minimal normal.
Now $1 < K < G$ is a chief series of $G$, so a minimal normal $N$ subgroup of $G$ is 


*

*either isomorphic to $K$, or

*of order 2,


If $N \ne K$, the first case cannot occur, since two distinct minimal normal subgroup have trivial intersection.
If $N$ has order 2, then $G \cong K \times N$, which is not the case.
