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 wonder if there is a nontrivial decomposition of the $L_2(q)$, where $q$ is a prime power, into a direct product. I think that there is none, but I am not sure.

$L_2(q)$ refers to the special projective group also denoted by $PSL(2,q)$.

Does the following argument holds?

Can one argue by considering first the general linear group $GL(2,q)$ and proving that it does not decompose into a direct product, and hence since $L_2(q)$ is the quotient of $SL(2,q)$ by its subgroup of scalar transformations with unit determinant, we can't have a decomposition for $L_2(q)$.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

I would suggest the following straightforward proof for $q > 3$.

It is well known (from the Classification of the Finite Simple Groups) that the groups $L_2(q)$ are simple for $q >3$. Suppose that for a given $q>3$, you can write $L_2(q) = A \times B$ for $A$ and $B$ nontrivial. Then $A$ and $B$ are normal in $L_2(q)$, a contradiction.

share|improve this answer
    
$L_2(2), L_2(3)$ are not simple. –  Qiaochu Yuan May 24 '11 at 13:16
    
You're right. I fixed it. –  Thomas Connor May 24 '11 at 13:21
    
Using the classification of finite simple groups to prove that the groups ${\rm L}_2(q)$ are simple is a serious case of using an ICBM to shoot a sparrow! (as well as being circular reasoning) –  Derek Holt May 25 '11 at 8:23

Your argument is not valid: it is not true that the quotient of a group which is not a nontrivial direct product is not a nontrivial direct product. For example, the quaternion group $Q_8$ is not a nontrivial direct product (since otherwise it would be abelian), but it has a quotient isomorphic to $C_2 \times C_2$.

But the result you want is true. $L_2(2) \cong S_3$ is not a nontrivial direct product (again, since otherwise it would be abelian), and neither is $L_2(3) \cong A_4$ (its only nontrivial normal subgroup is $C_2 \times C_2$). For all higher values of $q$ it is well-known that $L_2(q)$ is simple, hence has no nontrivial normal subgroups.

I don't actually know how hard it is to prove that these groups are simple. I can prove directly that they're not direct products if $q$ is prime. First, observe that $\text{PSL}_2(\mathbb{F}_q)$ acts double transitively and faithfully on the projective line $\mathbb{P}^1(\mathbb{F}_q)$. It follows that the representation corresponding to this permutation representation decomposes as the direct sum of the trivial representation and an irreducible faithful representation $V$ of dimension $q$. If $\text{PSL}_2(\mathbb{F}_q) \cong G \times H$, then $V \cong V_G \otimes V_H$ where $V_G, V_H$ are irreducible faithful representations of $G, H$.

Since $q$ is prime, it follows WLOG that $\dim V_G = p, \dim V_H = 1$, hence that $H$ is cyclic. Since $H$ is normal, it must be central, but $\text{PSL}_2(\mathbb{F}_q)$ has no center; contradiction.

share|improve this answer
    
Thanks a lot for this clear answer! –  Julien Meyer May 24 '11 at 13:55

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.