Prove that $G=$SL$(2, \mathbb{F}_5)$ is an extension of $\mathbb{Z}_2$ by $A_5$ which is not a semidirect product. Question: 

Prove that $G=$SL$(2, \mathbb{F}_5)$ is an extension of $\mathbb{Z}_2$ by $A_5$ which is not a semidirect product.

(This is a question from Rotman's Advanced Modern Algebra which I am trying to self-learn.)
I guess that we can send $\mathbb{Z}_2$ to
$$K = \{ 
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
,
\begin{bmatrix}
-1 & 0 \\
0 & -1 \\
\end{bmatrix}
\} $$
which is normal in $G$. and then I read online that $G/K$ is isomorphic to $PSL(2, \mathbb{F}_5)$ which is isomorphic to $A_5$.
So I can create the short exact sequence
$$ 0 \to \mathbb{Z}_2 \to^i G \to^p A_5 \to 1$$
Does this suffice to show that $G$ is an extension of $\mathbb{Z}_2$ by $A_5$?
Next, I want to show that this is not a semidirect product, which means I need to show that the extension is not split, which means I need to show there does not exist any homomorphism $j :A_5 \to G$ such that $pj$ is the identity on $A_5$. But how do I do this? Is it not possible to use the isomorphism $G/K \cong A_5$ to construct some homomorphism?
 A: Your work is totally fine.
Now, for your last question: if the sequence were split, then $A_5$ would be isomorphic to a subgroup of index $2$ of $G$, and you would get a surjective morphism $G\to \mathbb{Z}/2\mathbb{Z}$ with kernel isomorphic to $A_5$.
Now since $\mathbb{Z}/2\mathbb{Z}$ is abelian, this morphism sends any element of $[G,G]$  to $0$ (because any commutator is mapped to $0$. It is a well-known fact that $[SL_n(K),SL_n(K)]=SL_n(K)$ except if $n=2$ and $K=\mathbb{F}_2$ or $\mathbb{F}_3$. Using this classical fact, you see that your morphism is trivial, and you get a contradiction.
A: For the last part, note that by looking at Rational or Jordan Canonical forms one can easily see that $SL(2,5)$ has only one class of involutions, namely the class containing $-I$. This element is central, so there is exactly one involution. 
Hence the extension of the Sylow-2-subgroup of $SL(2,5)$ (or order $8$) by the central involution can't split, so neither can the extension of the whole group.   
