TITLE: Semi-direct products
QUESTION [8 upvotes]: I need some help to show that $\text{SL}_{2}(\mathbb{Z}_{3})$ is a semi-direct product of the quaternion group and the cyclic group $C_{3}$. Any ideas? I don't want to write the elements of $\text{SL}_{2}(\mathbb{Z}_{3})$, it seems like a very ugly method.
thanks

REPLY [5 votes]: An upper triangular matrix with all non-zero entries equal to $1$ and a lower triangular matrix of same type generate two cyclic subgroups of order $3$ - which are Sylow-3 subgroups. The number of Sylow-3 subgroups in $G=SL(2,3)$ is either $1$ or $4$; but by the remark, it should be $4$; say $P_1, P_2, P_3, P_4$. Hence index of normalizer of a Sylow-3 subgroup is $4$ : $[G:N(P_1)]=4$ i.e. $G$ has a subgroup of index $4$. (note that then $|N(P_1)|=6$)
Then there will be a homomorphism $\phi\colon G\rightarrow S_4$, with $ker(\phi)\subseteq N(P_1)$ (Generalised Cayley's Theorem).

If $ker(\phi)=N(P_1)$, then it will be a normal subgroup of $G$, and $P_1$ is a characteristic subgroup of $N(P_1)$ (since it is unique subgroup of order 3); hence $P_1\triangleleft G$, contradiction.
If $ker(\phi)=P_1$, then $P_1\triangleleft G$, contradiction
If $ker(\phi)=\{1\}$, then $G\cong S_4$ contradiction (why?)
Therefore, $|\ker(\phi)|=2$, and so $ker(\phi) \cong \mathbb{Z}_2$.
Then $G/ker(\phi)$ will be isomorphic to a subgroup of order 12 in $S_4$; we must have $G/\mathbb{Z}_2 \cong A_4$.

Since $A_4$ has a normal subgroup of index 3, by correspondence theorem, $G$ has a normal subgroup of index $3$, containing $ker(\phi)=\mathbb{Z}_2$. This normal subgroup of index $3$ (hence order 8) will be Sylow-2 subgroup of $G$.
Since $G$ has unique element of order $2$; the normal Sylow-2 subgroup will also have unique element of order 2, it is either $\mathbb{Z}_8$ or $\mathbb{Q}_8$. It can not be $\mathbb{Z}_8$ because the normal subgroup of index $3$ in $A_4$ contains 3 subgroups of index $2$; the corresponding subgroup $\mathbb{Z}_8$ should also contain three subgroups of index $2$, contradiction.
We have shown 

$G$ has normal Sylow-2 subgroup isomorphic to $\mathbb{Q}_8$

The rest part is then easy to answer the question:
Since $Q_8 \triangleleft G$, $\mathbb{Z}_3\leq G$, so $Q_8\mathbb{Z}_3\leq G$, and computing order we see $Q_8\mathbb{Z}_3=G$.
Therefore $Q_8\triangleleft G$, $\mathbb{Z}_3\leq G$, $G=Q_8\mathbb{Z}_3$, and $\mathbb{Z}_3 \cap Q_8=\{1\} \Rightarrow G=Q_8\rtimes \mathbb{Z}_3$