Lemma 3

Under the hypotheses of Lemma 1, assume that the $G_{\alpha}$ are simply-connected compact Lie groups. Then, the topological group G is isomorphic to the product of a family of almost simple, simply-connected compact Lie groups.

For all $\alpha \in I$, the group $G_{\alpha}$ is the direct product of a finite family of almost simple, simply-connected subgroups $(S^{\lambda}_{\alpha})_{\lambda\in L_{\alpha}}$ (Chap. III, §9, no. 8, Prop. 28). Let $\beta \in I,\beta \geq \alpha$. By Lemma 2, there exists a map $l_{\beta \alpha}: L_{\alpha}\rightarrow L_{\beta}$ such that $f_{\alpha \beta}(S^{l_{\beta \alpha}(\lambda)}_{\beta}) = S^{\lambda}_{\alpha}$ for $\lambda \in L_{\alpha}$. We have $l_{\gamma \beta}\circ l_{\beta \alpha}=l_{\gamma \alpha}$ for $\alpha \leq \beta \leq \gamma$, so $(L_{\alpha}, l_{\beta \alpha})$ is an inductive system of sets relative to I. Let L be its limit; the maps $l_{\beta \alpha}$ being injective, $L_{\alpha}$ can be identified with a subset of L, so that $L =\bigcup_{\alpha\in I}L_{\alpha}$.

Let $\lambda \in L$. Put $S^{\lambda}_{\alpha}=\{1\}$ when $\lambda \notin L_{\alpha}$, and denote by $\varphi^{\lambda}_{\alpha \beta}: S^{\lambda}_{\beta}\rightarrow S^{\lambda}_{\alpha}$ the morphism induced by $f_{\alpha \beta}$; this gives a projective system of topological groups $(S^{\lambda}_{\alpha}, \varphi^{\lambda}_{\alpha \beta})$, whose limit is isomorphic to $S_{\lambda}$. The canonical homomorphism of topological groups

lim$_{_{\leftarrow -\alpha\in I}}((\prod_{\lambda\in L}S^{\lambda}_{\alpha})\rightarrow \prod_{\lambda\in L}($lim$_{_{\alpha\leftarrow -\in I}}S^{\lambda}_{\alpha})$

is bijective (Theory of Sets, Chap. III, §7, no. 3, Cor. 2); it is thus an isomorphism since the groups in question are compact. But the first of these groups can be identified with G and the second with the product of the $S_{\lambda}$, hence the lemma.