# Exact sequence from $G=G_0\supset G_1\supset G_2\supset\cdots\supset G_r=\{e\}$

This is Exercise 5.3 from Algebra: Chapter 0.

Given a normal series of subgroups $$G=G_0\supset G_1\supset G_2\supset\cdots\supset G_r=\{e\},$$ construct an exact sequence of groups using $\{e\}, G$ and $H_j=G_j/G_{j+1}$ to connect $\{e\}$ and $G$.

I am not sure what the author is asking for. Starting from $\{e\}$, I think the only natural sequence involving only $\{e\}, H_j$ and $G$ is of the form $$\{e\}\longrightarrow G_{r-1}/G_{r}\longrightarrow G_{r-2}/G_{r-1}\longrightarrow\cdots.$$ However this is not exact.

Can someone give a hint? Thanks!

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Show how to 'connect' $\{e\}$ to $G$ by means of $r$ exact sequences of groups, involving only $\{e\}$, $G$ and the quotients $H_i = G_i/G_{i+1}$.
So you are not supposed to find a single exact sequence connecting $\{e\}$ to $G$, but rather $r$ seperate ones. I think the exercise in that form still contains a typo and should read "involving only $\{e\}$, $G_i$ and the quotients $H_i = G_i/G_{i+1}$". Then you can take (writing $1$ for $\{e\}$) \begin{matrix} 1 & \to & G_1 & \to & G & \to & H_0 & \to & 1,\\ 1 & \to & G_2 & \to & G_1 & \to & H_1 & \to & 1,\\ & & & & \vdots & & & & \\ 1 & \to & G_{r-1} & \to & G_{r-2} & \to & H_{r-2} & \to & 1,\\ 1 & \to & 1 & \to & G_{r-1} & \to & H_{r-1} & \to & 1. \end{matrix}