Introduction

Let $\mathbb{F}_3$ be the field of three elements and $SL_2(\mathbb{F}_3)$ be the group of $2 \times 2$ matrices with determinant $1$. In this post we show that $SL_2(\mathbb{F}_3)$ is the semi-direct product of $H_8$ and $\mathbb{Z}/3\mathbb{Z}$.

First of all we determine the cardinality of $SL_2(\mathbb{F}_3)$. To do this, we consider $GL_2(\mathbb{F}_3)$ and notice that $SL_2(\mathbb{F}_3)$ is the kernel of $\det$ function.

To determine $GL_2(\mathbb{F}_3)$, fix a basis of $\mathbb{F}_3 \oplus \mathbb{F}_3$ and let $A$ be a matrix representation of an element in $GL_2(\mathbb{F}_3)$. The first column of $A$ has $3^2-1$ number of choices: we only exclude the case of $(0,0)^T$. The second column has $3^2-3$ choices. We exclude $3$ possibilities given by the scalar multiplication of the first column to prevent linear dependence. Therefore $|GL_2(\mathbb{F}_3)|=(3^2-1)(3^2-3)=48$. Next we consider the exact sequence

We get $|SL_2(\mathbb{F}_3)|=|GL_2(\mathbb{F}_3)|/(\mathbb{F}_3)^\ast|=48/2=24$.

We immediately think about the possibility that $SL_2(\mathbb{F}_3)\cong \mathfrak{S}_4$. Is that the case?

24=3*8

As a group of order 24, we immediately consider the elements of order $2$, $3$ and $4$ in order to know the structure of the group we are looking at.

The element of order 2

There are ${4 \choose 2}/2!=3$ elements of order $2$ in $\mathfrak{A}_4$, i.e. those being products of two $2$-cycles. However, how many elements of order $2$ are there in $SL_2(\mathbb{F}_3)$? Let $A$ be such an element, then $A^2 = I$. Therefore all elements of order $2$ is nullified by the polynomial

If $A \in SL_2(\mathbb{F}_3)$ is of order $2$, then the minimal polynomial of $A$ divides $f(X)$, hence is either $X+1$ or $X^2-1$. The second case is impossible because then $f(X)$ will be the characteristic polynomial of $A$ and therefore $A$ has eigenvalue $1$ and $-1$ thus determinant $-1$. We get

Proposition 1. The element in $SL_2(\mathbb{F}_3)$ of order $2$ is only $A=-I$. In particular, $SL_2(\mathbb{F}_3)$ is not isomorphic to $\mathfrak{S}_4$.

Determine the group using Sylow theory

Checking elements of order $2$ is not out of nowhere. Since $24=2^3 \cdot 3$, it makes sense to look at $2$-Sylow and $3$-Sylow subgroups of $SL_2(\mathbb{F}_3)$. Sylow’s theorem ensures that there is a subgroup of order $3$, which can only be $\mathbb{Z}/3\mathbb{Z}$. We have also determined that the subgroup of order $2$ is $\{-I,I\}$. Next we determine the group of order $8$.

Elements of order 4

To study elements of order $4$, we immediately consider the polynomial

Let $A \in SL_2(\mathbb{F}_3)$ be an element of order $4$. Then $g(A)=0$. But since $A+I \ne 0$ and $A-I \ne 0$, we will be considering $h(X)=X^2+1$ instead. Notice that $h(X)$ is irreducible in $\mathbb{F}_3[X]$ and therefore it is minimal polynomial of $A$. Since the degree of $h$ is $2$, we also see $h(X)$ is the characteristic polynomial of $A$.

From this polynomial we see that $\mathrm{tr}(A)=0$. Combining with the fact that $|A|=1$, we can easily deduce that elements of order $4$ consists of

We in particular have $i^3=i^{-1}=-i$, $j^3=j^{-1}=-j$ and $k^3=k^{-1}=-k$. Furthermore, $k=ij=-ji$. These identities rings a bell of quaternion number. We therefore have the quaternion group lying in $SL_2(\mathbb{F}_3)$ as a $2$-Sylow subgroup:

Is there any other $2$-Sylow subgroup? The answer is no. To see this, let $H’$ be another $2$-Sylow group. Then there exists some $g \in SL_2(\mathbb{F}_3)$ such that $H’=gH_8 g^{-1}$, which is equal to $H_8$ because all elements in $K$ will have order $4$.

Proposition 2. The quaternion group $H_8$ can be embedded into $SL_2(\mathbb{F}_3)$ as the unique $2$-Sylow group. In particular, $SL_2(\mathbb{F}_3)$ has no element of order $8$.

An element of order 3

Let $A \in SL_2(\mathbb{F}_3)$ be an element of order $3$. Then its minimal polynomial $m(X)$ divides $X^3-1=(X-1)^3=(X-1)^2(X-1)$. Since $A-I \ne 0$, we must have $m(X)=(X-1)^2=X^2+X+1$. We can also see that the characteristic polynomial of $A$ is also $X^2+X+1$. In particular, we see the trace of $A$ is $-1=2$. We can then choose

Therefore $K=\{I,A,A^2\}$ is a $3$-Sylow subgroup of $SL_2(\mathbb{F}_3)$, which is not unique, because for example one can also consider the group generated by the transpose of $A$.

Conclusion

Notice that $H \cap K = \{1\}$ because $\gcd(3,4)=1$. Therefore the map $H \times K \to HK$ given by $(x,y) \mapsto xy$ is bijective. Since $H$ is also normal, we are safe to write $G=H\ltimes K$ because $|HK|=|H||K|=24=|G|$.

References

• Serge Lang, Algebra Revised Third Edition
• Olivier SERMAN, $SL_2(\mathbb{F}_3)$.