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Let $G = \hbox{proj.lim.}_{\alpha} \{ G_{\alpha} , \varphi^{\alpha}_{\beta} \}$ be a projective limit of simple groups (i.e., each $\varphi^{\alpha}_{\beta}\colon G_{\alpha}\to G_{\beta}$ is a surjective group homomorphism between simple groups).

It is clear that if $\varphi^{\alpha}_{\beta}$ is not the trivial map ($x\mapsto 1$), then $\varphi^{\alpha}_{\beta}$ is an isomorphism. I believe $G$ is always a simple group. But I don't know how to prove this.

Any suggestions?

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    Ehr... first, there are more conditions to being a "projective limit" in general; is your index set ordered? If so, you also need that if $\alpha\leq\beta\leq\gamma$, then $\varphi_{\alpha}^{\gamma}=\varphi_{\alpha}^{\beta}\varphi_{\beta}^{\gamma}$. And the only way the map is not an isomorphism is if $G_{\beta}$ is trivial. So either all your maps are trivial, or from some point "back" they are all isomorphisms so $G$ is just yet another isomorphic copy of $G$.2010-11-12
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    Arturo's comment provides a succinct form of the argument I give below. (My answer may still be useful as a detailed guide to making these sorts of arguments. Also, I do assume the indexing set is directed, and that the transition maps satisfy the compatibility condition that Arturo mentions.)2010-11-12
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    I think in the definition of an inverse system is not assumed $A$ is directed.2010-11-13

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