Exercise 8.1.4

Proof. As ${\tilde{G}}_i◃{\tilde{G}}_{i-1}$, and as $\pi$ is a group homomorphism, then ${\pi }^{-1}({\tilde{G}}_i)◃{\pi }^{-1}({\tilde{G}}_{i-1})$, so $G_i◃G_{i-1}$.

Indeed, if $x\in G_{i-1},y\in G_i$, then $x^{\prime }=\pi (x)\in {\tilde{G}}_i,y^{\prime }=\pi (y)\in {\tilde{G}}_{i-1}$, where ${\tilde{G}}_i◃{\tilde{G}}_{i-1}$, so $x^{\prime }{y}^{\prime }{x{}^{\prime }}^{-1}\in {\tilde{G}}_i$, then $\pi (\mathit{xy}{x}^{-1})=\pi (x)\pi (y)\pi {(x)}^{-1}\in {\tilde{G}}_i$, and $\mathit{xy}{x}^{-1}\in {\pi }^{-1}({\tilde{G}}_i)=G_i$.

As $\pi$ est surjective, $\pi (G_i)={\tilde{G}}_i$, and the situation is the same as in Exercise 2, where we have proved that $\bar{\phi }:G_{i-1}∕G_i\to {\tilde{G}}_{i-1}∕{\tilde{G}}_i$ given by $g{G}_i\mapsto \pi (g){\tilde{G}}_i$ is well defined, and is a surjective group homomorphism. It remains to verify that $\bar{\phi }$ is injective.

If $g{G}_i\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\bar{\phi })$, then $\bar{\phi }(g{G}_i)={\tilde{G}}_i$, that is $\pi (g){\tilde{G}}_i={\tilde{G}}_i$, thus $\pi (g)\in {\tilde{G}}_i$, so $g\in G_i$, and $g{G}_i=G_i$ is the identity of $G∕G_i$. Therefore $\bar{\phi }$ is injective. $\bar{\phi }$ is a group isomorphism:

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