Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am asking kindly,

For which values of $n$ we have $$S_n≅P\Gamma L_2(3),S_n≅P\Gamma L_2(4)$$ This may be correct if we replace $S_n$ by $A_n$.

Any help will be appreciated. :)


Edit (JL): Adding the definition of the group. Let $\Omega$ denote the set $GF(q)\cup\{\infty\}$. Then the group $P\Gamma L_2(q)$ consists of bijections from $\Omega$ to itself the type $$ P\Gamma L_2(q)=\{f:\Omega\rightarrow\Omega\mid f(z)=\frac{az^\sigma+b}{cz^\sigma+d}; a,b,c,d\in GF(q); ad-bc\neq0; \sigma\in Aut(GF(q))\}. $$

This group has order $q(q^2-1)\cdot |Aut(GF(q))|$.

share|improve this question
1  
What does $P\Gamma L_2$ mean? Like, $PSL(\Bbb F_3^2)$? Every finite group is isomorphic to a subgroup of a symmetric group... –  anon Jul 1 '12 at 6:24
    
$P\Gamma L_2(q)=\{f|f:\Omega\longrightarrow\Omega, f(z)=\frac{az^\sigma+b}{cz^\sigma+d},ad-bc\neq 0; a,b,c,d\in GF(q)\}$, $\Omega=GF(q)\cup\{\infty\}$ and $\sigma\in Aut(GF(q))$. Yes exactly, but for some unknown values these groups $P\Gamma L_2(3)$ or $P\Gamma L_2(4)$ becomes isomorphic with some copy of $S_n$ or even $A_n$. I want to know these $n$. –  B. S. Jul 1 '12 at 6:33
    
Wait, isomorphic to a permutation group (subgroup of an $S_n$), or isomorphic to a symmetric or alternating group ($S_n$ or $A_n$ for some $n$)? These are radically different questions. –  anon Jul 1 '12 at 6:40
    
@anon: Maybe I wrote my answer incorrectly, but I want to find these $n$. I have no refrences noting for what value of $n$, the kenrel of the actions would be $\{\}$. Thanks for your time. –  B. S. Jul 1 '12 at 6:48
1  
Well, if that is the group order, then with $q=3$ we get a group of order $3\cdot 8\cdot1=24$, because the Galois group is trivial. Is this the order of a symmetric group $S_n$ or the alternating group $A_n$ for some $n$? Once you figure that out, then you know which isomorphism you need to prove. –  Jyrki Lahtonen Jul 1 '12 at 7:19

1 Answer 1

up vote 3 down vote accepted

By definition $P\Gamma L_2(q)$ is a subgroup of $\mathrm{Sym}(\Omega)\simeq S_{q+1}$. The claims would follow immediately from the knowledge of the order of the group $P\Gamma L_2(q)$:

When $q=3$, the group $Aut(GF(3))$ is trivial, and we have $|P\Gamma L_2(3)|=24=|S_4|$, so it follows that $P\Gamma L_2(q)$ must be all of $S_4$.

When $q=4$, the group $Aut(GF(4))$ is cyclic of order tow, and we have $$|P\Gamma L_2(4)|=4\cdot(4^2-1)\cdot2=120=|S_5|.$$ Again we see that $P\Gamma L_2(4)$ must be all of $S_5$.

So the real question is to prove the formula for the order of the group in these two cases.

In the case $q=3$ we observe that the transformations $z\mapsto z+1$ and $z\mapsto 2z$ give the permutations $\alpha=(012)(\infty)$ and $\beta=(12)(0)(\infty)$ of $\Omega$ respectively. Clearly these two generate the group $S_3$ acting on the affine part $GF(3)$. The transformation $z\mapsto 1/z$ gives the permutation $\gamma=(0\infty)(1)(2)$. Together with $\alpha$ we get a group that acts transitively on $\Omega$, because $\gamma(\infty)=0$, and $(\alpha^i\gamma)(\infty)=i$ for $i=1,2$. Therefore the group generated by all of $\alpha,\beta,\gamma$ must have at least $4\cdot|\langle \alpha,\beta\rangle|=4\cdot6=24$ elements, and hence be all of $S_4$.

For the case $q=4$ I write $GF(4)=\{0,1,a,b=a+1=a^2\}$. Here $F:x\mapsto x^2$ is the only non-trivial automorphism. It corresponds to the permutation $\phi=(0)(1)(ab)(\infty)$. In addition to that we have $z\mapsto z+1$ giving us $\alpha=(01)(ab)(\infty)$, and $z\mapsto az$ giving us the permutation $\beta=(0)(1ab)(\infty)$. The elements $\phi$ and $\beta$ fix both $0$ and $\infty$, and we see that they generate a copy of $S_3$ acting on $\{1,a,b\}$. The group of translations $K$ generated by $\alpha$ and $\beta\alpha\beta^{-1}$, the latter corresponding to $z\mapsto z+a$, gives us a copy of the Klein four group acting transitively on the finite part $GF(4)$. As in the preceding case we thus see that the group $H=\langle \alpha,\beta,\phi\rangle$ generated by these three transformations must act as $S_4$ on $GF(4)=\Omega\setminus\{\infty\}$ because it has at least $4\cdot6=24$ elements. It remains to show that adding the generator $\gamma=(0\infty)(1)(ab)$ gives us a group of size at least $120$. This follows from transitivity as in the case $q=3$. We can map $\infty$ to $0$ with $\gamma$, so $\gamma$ followed by an appropriate element of $K$ maps $\infty$ to any other element of $\Omega$. Therefore the order of the group $G=P\Gamma L_2(4)$ is at least $5\cdot|H|=120$. As $G$ was known to be a subgroup of $S_5$, the isomorphism $G\simeq S_5$ follows.

share|improve this answer
    
Thanks for helping me step by step. By the way, we need a new tag here, "Linear Groups". I cannot make it but if think it is necessary to have this new tag, please make it for MSE. :-) –  B. S. Jul 1 '12 at 10:53
    
@Babak: Hmm. I'm undecided about the need of a new tag. I would use one of algebraic-groups, lie-groups or finite-groups all depending. It may be that there would be a benefit of singling out finite linear groups, so I'm not saying "No", either. I have never created a new tag, so I'm cautious. You can ask about this in meta. –  Jyrki Lahtonen Jul 1 '12 at 11:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.