Group homomorphism from $SL_2 (\mathbb Z / 5 \mathbb Z)$ to $S_5$ I need to find a group homomorphism from $SL_2 (\mathbb Z / 5 \mathbb Z)$ to $S_5$. There is obviously the trivial homomorphism but I am then asked to deduce that $SL_2 (\mathbb Z / 5 \mathbb Z)/\{I,-I\} \cong A_5$, so ideally I want a homomorphism with kernel $\{I,-I\}$, and image $A_5$ so I can use the isomorphism theorem. Any hints?
 A: 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.
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A: Just wanted to add some extra commentary.

Let $I$ be the rotational symmetry group of the (regular) icosahedron. By orbit-stabilizer, it has order $60$. It is a subgroup of $\mathrm{SO}(3)$, the full group of $3\times3$ rotation matrices. Every edge of the icosahedron uniquely determines a compound of three inscribed semiperpendicular golden rectangles:
$\hskip 0.5in$ 
There are $6$ icosahedron edges per compound, so there are $30/6=5$ compounds. Note $I$ acts on this set $\Omega$ of $5$ compounds, indeed we may identify $I$ with an index $2$ subgroup of $\mathrm{Sym}\,\Omega$, forcing it to be the alternating subgroup by uniqueness. Thus, $I\cong A_5$ by labelling the compounds $1$ thru $5$.
On the other hand, there is a double cover $2I$ of the icosahedral group (called the binary icosahedral group), which is the preimage of $I$ under the $2$-to-$1$ group homomorphism $S^3\to\mathrm{SO}(3)$, where $S^3$ is the group of versors (unit quaternions). (There are many resources discussing how the antipodal versors $\pm \exp(\theta\mathbf{u})$ represent the $3$D rotation around the $\mathbf{u}$-axis by $2\theta$, where $\mathbf{u}$ is a unit vector.) By considering quaternions as a $2$D complex vector space, we may identify quaternions with $2\times2$ complex matrices, in which case $S^3$ may be identified with $\mathrm{SU}(2)$. Then the elements of $2I$ may be reduced mod $5$ to get $\overline{2I}=\mathrm{SL}_2(\mathbb{F}_5)$. Since the kernel of $2I\to I$ corresponds to the kernel of $\mathrm{SL}_2(\mathbb{F}_5)\to\mathrm{PSL}_2(\mathbb{F}_5)$, we conclude $I\cong\mathrm{PSL}_2(\mathbb{F}_5)$.
Thus, $\mathrm{PSL}_2(\mathbb{F}_5)\cong A_5$. I explained this in my answer here.

Symmetric groups have trivial outer automorphism group, except for the exceptional case of $\mathrm{Out}\,S_6=\mathbb{Z}_2$. An outer automorphism of $S_6$ swaps cycle types $(\cdot\cdot)$ and $(\cdot\cdot)(\cdot\cdot)(\cdot\cdot)$ as well as $(\cdot\cdot\cdot)$ with $(\cdot\cdot\cdot)(\cdot\cdot\cdot)$. While the subgroup $A_5$ stabilizes a point, its image under an outer automorphism is an "exotic" copy of $A_5$ which is actually transitive. This matches (up to relabelling) the action of $\mathrm{PSL}_2(\mathbb{F}_5)$ on $\mathbb{F}_5\mathbb{P}^1$ and of $I$ on the set $\Phi$ of antipodal pairs of vertices of the icosahedron (note $|\mathbb{F}_5\mathbb{P}^1|=|\Phi|=6$ in contrast to $|\Omega|=5$).
