projective limit of simple groups 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?
 A: 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.
