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Quotient of Quotient Group

Writer Matthew Harrington
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Say I have $N_1$ a normal subgroup of $G$, $N_2$ a normal subgroup of $G/N_1$, $N_3$ a normal subgroup of $((G/N_1)/N_2)$, ... $N_r$ normal subgroup of $((((G/N_1)/N_2)/N_3)/ ... )/N_{r-1}$. What does the identity element of $((((G/N_1)/N_2)/N_3)/ ... )/N_r$ look like in coset notation of G? What does a general element of that group look like? I'm looking for an analogy to $G/N_1 = \{gN_1 | g \in G\}$.

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2 Answers

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The identity is $e\cdot N_1\cdot....\cdot N_k$ and a general element looks like $g\cdot N_1\cdot...\cdot N_k$.

Warning: Since $N_1,N_2,...,N_k$ are subgroups of the different groups one can't talk about their product and therefore it is important to read the terms above with the brackets. It should be:

$$(...(((gN_1)N_2)N_3)...N_k)$$

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As a supplement to Yanko's answer, note that an element of $(\ldots((G/N_1)/N_2)/N_3 \ldots /N_{r-1})/N_r$ is an element of the $r$-fold iterated power set, $\Bbb{P^r}G$. I.e., it is a set of sets of sets of ... of subsets of $G$ with $r+1$ "of"s. If $$\pi_i: (\ldots((G/N_1)/N_2)/N_3 \ldots /N_{i-2})/N_{i-1} \to (\ldots((G/N_1)/N_2)/N_3 \ldots /N_{i-1})/N_i$$ is the natural projection, $i = 1, \ldots r$, then by $r$ applications of the third isomorphism theorem $(\pi_{r} \circ \cdots \circ \pi_1)^{-1}[N_r]$ is a normal subgroup $K_r$ say of $G$ containing $N_1$ and $(\ldots((G/N_1)/N_2)/N_3 \ldots /N_{r-1})/N_r$ is naturally isomorphic to $G/K_r$ (whose elements are easier to represent, if you really need to know what the representation is, which is rare).

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