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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

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Look at $SL_2(\mathbb{Z}_3)$. Find a copy of $Q_8$, $H$ say, and a copy of $C_3$, $K$ say. Do they intersect trivially? Is one normal? Does $G=KH$? –  user1729 Oct 10 '11 at 13:55
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I think the easiest way is just to check the elements. Here are two ways to prove it, but they use more group theory.

2-local: If q is (a field whose order is) an odd prime power, then SL2(q) has quaternion Sylow 2-subgroups H generated in one of two ways depending on q mod 4. Check that in fact the standard Sylow q-subgroup K (consisting of upper-triangular matrices with 1s on the diagonal) normalizes H when q = 3, and that by order considerations the resulting semi-direct product is the entire group.

3-local: If q is (a field whose order is) a prime power, then SLn(q) has Sylow q-subgroups consisting of upper-triangular matrices with 1s on the diagonal, and its normalizer is all upper-triangular matrices (with determinant 1). However, when n = 2 and q = 3, the only possible diagonals are [1,1] and [−1,−1], both of which centralize the Sylow q-subgroup. Hence the Sylow q-subgroup has a normal complement, by Burnside, and in this case that complement is the entire Sylow 2-subgroup, which is quaternion as usual since n = 2 and q is odd.

Maybe though you are just asking how to remember which direction the semi-direct product goes. I always forget, even though it should be important enough for me to remember. I developed these two ways to remember:

Normalizer mnemonic: SL2(odd) generally has quaternion Sylow 2-subgroups; call one H. In this group of order 24, the normalizer is either all of SL2(q) and we are done, or the normalizer is just H itself. In the latter case, we'd have that K was normal by Burnside and the fact that proper subgroups of Q8 have 2-groups as automorphism groups. So either you have C3 acting on Q8 or Q8 acting on C3. Both semi-direct products occur, but only one of them is SL2(3); how do you remember which one?

Well, SL2(K) is usually a perfect group; it usually has no abelian quotients. The only exceptions to these matrix group identities for finite fields are always in the defining characteristic. SL2(2) has an abelian quotient of order 2, and SL2(3) has an abelian quotient of order 3. Hence it is the C3 acting on Q8.

Quotient mnemonic The other mnemonic is that PSL(2,3) = Alt(4) has a normal Sylow 2-subgroup, and Z(SL(2,3)) has order 2, so SL(2,3) has a normal Sylow 2-subgroup (which is quaternion as usual). This fits in with the other exceptions PSL(2,4) = PSL(2,5) = Alt(5) and PSL(2,9) = Alt(6).

I'd be interested in a better mnemonic:

Is there some natural matrixy thing that SL2(3) acts on as a 3-cycle?

Is there some general reason SL2(q) would be acting on a vector space over the field with 2-elements?

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Jack, since $q=p^n$, when you say Sylow $q$-subgroup, it should be $p$-subgroup, right? –  Nathan Portland Oct 13 '11 at 22:04
    
@Nathan: yup, though I like thinking of them as Sylow q-subgroups since they still have order a power of q, and the same as structure as their p^1-friends if one allows more general exponents. –  Jack Schmidt Oct 13 '11 at 22:12
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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 $\mathbb{Z}_3 \cap Q_8=\{1\} \Rightarrow G=Q_8\rtimes \mathbb{Z}_3$

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