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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 $L_2(q)$.

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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.

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$L_2(2), L_2(3)$ 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.

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

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