# 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. 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. Jan 12 '15 at 23:41
• (It is interesting to note that ${\rm SL}(2,5)$ is not isomorphic to $S_5$, however!) Jan 12 '15 at 23:48
• This is related to math.stackexchange.com/questions/1096446 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. 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. 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. Jan 13 '15 at 15:23

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 $$|\Omega|=5$$).