Show that $G_i/G_{i+1} \twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$.

Question

From Aluffi's book of Algebra:

in which it refers to Theorem 7.12 :

[Red underlined:] I can't see any connection with Theorem 7.12! Actually I don't know how the $+$ appears. How to show that $G_i/G_{i+1} \twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$?

Answer 1

Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo. Let $G'=(G_iN)/(G_{i+1}N)$ and \begin{align} &\varphi:G_i\twoheadrightarrow G'& &x\mapsto xG_{i+1}N \end{align} Since $G_{i+1}\subseteq\operatorname{Ker}\varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $\bar\varphi:G_i/G_{i+1}\to G'$ making the following diagram commutative: $\require{AMScd}$ \begin{CD} G_i@>\varphi>>G'\\ @V\pi VV @|\\ G_i/G_{i+1}@>>\bar\varphi>G' \end{CD} where $\pi:G_i\to G_i/G_{i+1}$ is the canonical projection onto the factor group. Since $\varphi$ is surjective, $\bar\varphi$ is surjective as well.

If $G_i/G_{i+1} $ is simple, then $\operatorname{Ker}\bar\varphi=\{1\} $ or $\operatorname{Ker}\bar\varphi=G_i/G_{i+1} $, hence $\bar\varphi $ is an isomorphism or is trivial.