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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 $G$. – Arturo Magidin Nov 12 '10 at 4:01
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.) – Matt E Nov 12 '10 at 4:19
I think in the definition of an inverse system is not assumed $A$ is directed. – Immanuel Nov 13 '10 at 19:40

Suppose that for every $\beta$, there is $\alpha \geq \beta$ such that $\phi_{\beta}^{\alpha}$ is trivial. Then the projective limit is trivial. (Because if $(x_{\beta})$ is an element of the projective limit, then for each $\beta$, we have $x_{\beta} = \phi_{\beta}^{\alpha}(x_{\alpha}) = 1,$ if we choose $\alpha$ for the given $\beta$ so that $\phi_{\beta}^{\alpha}$ is trivial --- as we may, by assumption.)

On the other hand, suppose that for some $\beta$, the maps $\phi_{\beta}^{\alpha}$ are isomorphisms for all $\alpha \geq \beta$. If $A$ is the directed set indexing the projective limit, then replacing $A$ by its subset $A_{\geq \beta} := \{\alpha \in A \, | \ \alpha \geq \beta\}$ doesn't change the projective limit (since $A_{\geq \beta}$ is cofinal in $A$), and now all the transition maps $\phi_{\gamma}^{\alpha}$ for $\alpha, \gamma \in A_{\geq \beta}$ are isomorphisms. (Since each of $\phi_{\beta}^{\gamma}$ and $\phi_{\beta}^{\alpha} = \phi_{\beta}^{\gamma}\circ \phi_{\gamma}^{\alpha}$ is an isomorphism, by assumption.)

Now the projective limit of the $G_{\alpha}$ (for $\alpha \in A_{\geq \beta}$) has all its transition maps being isomorphisms, and so is isomorphic to $G_{\beta}$, and hence is simple.

Thus the projective limit is either trivial or simple, depending on whether one is in the first or second case.

Added in response to Arturo Magidin's comment below: I hadn't paid attention to the fact that the transition maps are assumed surjective. This implies that in my first case, all the $G_{\beta}$ are trivial (!), since they are (by assumption) the image of some $G_{\alpha}$ under the trivial map.

So unless all the $G_{\beta}$ are trivial, then one must be in the second case, when the $\phi^{\alpha}_{\beta}$ are all eventually isomorphisms.

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@Matt E: They've monkeyed around with the behavior of the renderer, and it no longer matches the previewer. You don't need the \\{ and \\} anymore, but you can't tell until after you post... – Arturo Magidin Nov 12 '10 at 4:17
@Arturo: Dear Arturo, Thanks for your comment. I just discovered this annoying phenomenon! – Matt E Nov 12 '10 at 4:18
@Matt E: Note that since one assumption is that all structure maps are surjective, the only way the first option can occur is if all $G_{\alpha}$ are themselves the trivial group. – Arturo Magidin Nov 12 '10 at 4:18
@Arturo: I hadn't paid attention to that. You are correct of course! I added a remark to my answer taking this into account. – Matt E Nov 12 '10 at 4:21
Are you assuming $A$ is directed? I can only follow the 3rd. paragraph of your answer under such assumption. – Immanuel Nov 13 '10 at 19:42

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