I had not seen elementary argument than the below. $PSL_2(5)$ can be considered as group of Mobius transformations
$$z\mapsto \frac{az+b}{cz+d}, (a,b,c,d\in\mathbb{Z}_5, ad-bc=1)$$
for $z\in \{0,1,2,3,4\}\cup\{\infty\}$, where $\{0,1,2,3,4\}=\mathbb{Z}_5$ and $\infty=1/0=2/0=...=4/0$, and some conventions similar to this.
The group $PSL_2(5)$ acts on these six symbols, whereas we want an action on $5$ symbols to show that it is isomorphic to $A_5$. So how to proceed.
The way I would like to proceedis to do the following:
In $PSL_2(5)$, find an element $A$ of order $2$, $B$ of order $3$ such that product $AB$ has order $5$.
This is quite easier: consider $A(z)=-1/z$. You see that it is of order $2$.
To get element of order $3$, note that if $B^3=I$ in $2\times 2$ matrix form then $B$ satisfies polynomial $t^3-1=(t-1)(t^2+t+1)$; the second factor is irreducible over $Z_5$; so take companion matrix $\begin{bmatrix} 0 & -1 \\ 1 & -1\end{bmatrix}$.
The corresponding Mobius map is $B(z)=(0.z-1)/(1.z-d)=-1/(z-1)$; check- this is of order $3$ (i.e. $B^3(z)=z$)
Finallt consider $AB(z)$; it is $AB(z)=z-1$, great! this is of order $5$ since this permutes $0,1,2,3,4$ in fashion of cycle of order $5$ and leaves $\infty$ fixed (by additive convention with $\infty$: $\infty+a=a+\infty=\infty$).
Thus, we found elements $A,B$ of order $2,3$ respectively such that $AB$ has order $5$. The group $A_5$ has presentation of this form:
$$A_5=\langle x_1,x_2: x_!^2=1, x_2^3=1, (x_1x_2)^5=1\rangle.$$
This gives a surjective homomorphism from $PSL_2(5)$ to $A_5$ ($A\mapsto x_1, B\mapsto x_2$).
Finally $PSL_2(5)$ and $A_5$ have same orders; the surjective homomorphism should be isomorphism.
Best Answer
There remain just 3 other subgroups which are order of 4 and because $|G:H|=8:4=2$, they are normal and $G/H=\mathbb Z/2$
So for each of them there is 3 cases and totally 9 cases,and overall $9+6=15$ homomorphisms
So for example, if $kerf=\{1,i,-1,-i\}$ ,
then cosets are $K,Kj\quad where \quad(Kj)^2=1$
Hence there is 3 cases depending on where $Kj$ goes
EDİT: I've forgotten the case $H=Q_8 $
So, total number is $16$