Group homomorphism from $SL_2 (\mathbb Z / 5 \mathbb Z)$ to $S_5$

I couldn't claim that this is an easiest way; but I have explained once this solutionway in my classroom for undergraduates.

The group $A_5$ has presentation $\langle x,y\colon x^2=1, y^3=1, (xy)^5=1\rangle$.

We can obtain a homomorphism you expect by this presentation. We find a pair of elements $X,Y$ in $SL(2,5)$ such that $X$ has order $2$, $Y$ has order $3$ and $(XY)$ has order $5$.

Each element of $SL(2,5)$ gives a Mobius transformation $z\mapsto \frac{az+b}{cz+d}$ where $a,b,c,d$ are in the domain=codomain=$\{0,1,2,3, 4,\infty\}$ of the transformation (convention $1/0=\infty$ and $1/\infty=0$). So we may try to find two Mobius transformation, one of order $2$, one of order $3$ so that their product has order $5$.

One choice for Mobius map of order $2$ is $z\mapsto \frac{-1}{z}$. For order $3$ Mobius map, one possibility is to find a map $z\mapsto \frac{az+b}{cz+d}$, such that $0\mapsto 1\mapsto \infty\mapsto 0$ (so that its order will be $3$), and it is in fact $z\mapsto \frac{1}{1-z}$.

Now it is easy to check that their product turns out to be order $5$.

What are the corresponding element in $SL(2,5)$? Well. These are $$X= \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} \mbox{ and } Y= \begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix}$$ which can be written just by considering the coefficients involved in the Mobius transformations. Now, the map $X\mapsto x$ and $Y\mapsto y$ can be uniquely extended to a homomorphism $f\colon SL(2,5)$ to $A_5$ (or $S_5$) (from the definition of the free group; I will not make it too technical.) And, thats it. .