# Is a primitive permutation group, indecomposable?

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No. There is a primitive action of $\operatorname{PSL}(2,7)\times\operatorname{PSL}(2,7)$ of degree $168$, that I found by computer search. – James Aug 30 '14 at 2:32
@James: You're right, with GAP PrimitiveGroup(168,1) is PSL(2, 7)^2. How did you find it? Is it the first? Did you write a function IsIndecomposable? – Sebastien Palcoux Aug 30 '14 at 9:03
It's not the first, because PrimitiveGroup(60,1) gives Alt(5)^2. Is it the first example? It seems that for a simple group $G$, the diagonal subgroup $D$ of $G×G$ is core-free and maximal (i.e primitive action of $G×G$ on the cosets $(G×G)/D$). Conversely, if $G \times G$ admits a primitive action, is $G$ simple? – Sebastien Palcoux Aug 30 '14 at 10:21
@SébastienPalcoux You should look up the O'Nan-Scott Theorem, which gives a precise description of all finite primitive permutation groups. In particular, for finite nonabelian simple group $S$, the action of $S$ on the cosets of a diagonal subgroup is a decomposable primitive group. I think these are probably the only examples, but I am not totally sure! – Derek Holt Aug 30 '14 at 11:25
The answer to your question, if $G \times G$ admits a primitive action then is $G$ (nonabelian) simple is yes. You just need to prove that any maximal subgroup of $G \times G$ contains a nontrivial normal subgroup of $G$. – Derek Holt Aug 30 '14 at 11:28