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?

• You could show that ${\rm PSL}(2,5)$ is a simple group of order $60$, and use thus that $A_5$ is the unique simple group of order $60$. You can show more generally that ${\rm PSL}(2,K)$ is simple provided $K$ has more than three elements. – Pedro Tamaroff Jan 12 '15 at 23:14
• As a general tactic, morphisms $G\to S_n$ arise whenever $G$ has a subgroup of index $n$, see this. If you could show that ${\rm SL}(2,5)$ has a subgroup of index $5$ that contains the center of ${\rm SL}(2,5)$ you might be in good path. – Pedro Tamaroff Jan 12 '15 at 23:41
• (It is interesting to note that ${\rm SL}(2,5)$ is not isomorphic to $S_5$, however!) – Pedro Tamaroff Jan 12 '15 at 23:48
• This is related to math.stackexchange.com/questions/1096446 – Derek Holt Jan 13 '15 at 9:19
• @PedroTamaroff Yes, there are three non-solvable groups of order $120$, $S_5$ with a $2$ at the top, ${\rm SL}(2,5)$ with a $2$ at the bottom, and $A_5 \times C_2$ with both. – Derek Holt Jan 13 '15 at 9:21

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. .

• You said that you were looking for $X$ and $Y$ in ${\rm SL}(2,5)$ such that $X$, $Y$, and $XY$ have orders $2,3,5$, but in your final choice of $X$ and $Y$, they have orders $4,5,3$. Note also that the only element of ${\rm SL}(2,5)$ of order $2$ is $-I_2$, which is in its centre. – Derek Holt Jan 13 '15 at 13:17
• Yaa! If we go modulo $\{\pm I\}$, then we get a homomorphism. So here $SL(2,5)\rightarrow PSL(2,5)\rightarrow A_5$ is the required homomorphism, where first one is quotient and second is obtained by $\bar{X}\mapsto x$, $\bar{Y}\mapsto y$. Thanks for notification. – Groups Jan 13 '15 at 15:23