Question

Let be a group and assume that for each positive integer i , ${\mathbf{N}}_{\mathbf{i}}$ is a normal subgroup of G. If every element of G can be written uniquely in the form ${\mathit{n}}_{{\mathbf{i}}_{\mathbf{1}}}\mathbf{‐}{\mathit{n}}_{{\mathbf{i}}_{\mathbf{2}}}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{n}}_{{\mathbf{i}}_{\mathbf{k}}}$with ${\mathbf{i}}_{\mathbf{1}}\mathbf{<}{\mathbf{i}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{<}{\mathbf{i}}_{\mathbf{k}}$and ${\mathbf{n}}_{{\mathbf{i}}_{\mathbf{j}}}\mathbf{\in }{\mathbf{N}}_{{\mathbf{i}}_{\mathbf{j}}}$, prove that $\mathbf{G}\mathbf{=}\underset{\mathbf{ieI}}{\mathbf{\sum }{\mathbf{N}}_{\mathbf{i}}}$(see Exercise 34). [Hint: Adapt the proof of Theorem 9.1 by defining$\mathbf{f}\left(a_1,a_2...\right)$ to be the product of those that are not the identity element.]

Short answer

It has been proved that,$\mathrm{G}\cong \underset{\mathrm{ieI}}{\sum {\mathrm{N}}_{\mathrm{i}}}$ .

Step-by-step solution

Given that

Given that $\mathrm{G}$is a group and for each positive integer i, ${\mathrm{N}}_{\mathrm{i}}$is a normal subgroup of $\mathrm{G}$. Every element of G can be written uniquely in the form${\mathrm{n}}_{{\mathrm{i}}_1}‐{\mathrm{n}}_{{\mathrm{i}}_2}...{\mathrm{n}}_{{\mathrm{i}}_{\mathrm{k}}}$ with${\mathrm{i}}_1<{\mathrm{i}}_2<...<{\mathrm{i}}_{\mathrm{k}}$ and ${\mathrm{n}}_{{\mathrm{i}}_{\mathrm{j}}}\in {\mathrm{N}}_{{\mathrm{i}}_{\mathrm{j}}}$.

Define a map which is surjective

As per the hint, define a map $\mathrm{f}:{\mathrm{N}}_1\mathrm{x}{\mathrm{N}}_2\mathrm{x}...\to \mathrm{G}\mathrm{by}\mathrm{f}\left({\mathrm{n}}_1,{\mathrm{n}}_2,...\right)={\mathrm{n}}_1{\mathrm{n}}_2...$

Since every element of G can be written in the form ${\mathrm{n}}_1{\mathrm{n}}_2..{.}_{}$and by the hypothesis, f is surjective.

Prove injective

$$\begin{gathered} \mathrm{Let}\mathrm{f}\left({\mathrm{n}}_1,{\mathrm{n}}_2,...\right)=\mathrm{f}\left({\mathrm{m}}_1,{\mathrm{m}}_2,...\right). \\ \mathrm{Then}{\mathrm{n}}_1{\mathrm{n}}_2...={\mathrm{m}}_1{\mathrm{m}}_2... \\ \mathrm{By}\mathrm{uniqueness}\mathrm{hypothesis},{\mathrm{n}}_{\mathrm{i}}={\mathrm{m}}_{\mathrm{i}}\mathrm{for}\mathrm{each}\mathrm{i}. \\ \mathrm{Therefore},\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)=\left({\mathrm{m}}_1{\mathrm{m}}_2...\right)\mathrm{in}{\mathrm{N}}_1\mathrm{x}{\mathrm{N}}_2\mathrm{x}... \\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{injective}. \end{gathered}$$

Prove homomorphism

It is clear that, N's are mutually disjoint subgroups and Lemma 9.2 implies that $$\begin{gathered} {\mathrm{n}}_{\mathrm{i}}{\mathrm{m}}_{\mathrm{j}}={\mathrm{m}}_{\mathrm{j}}{\mathrm{n}}_{\mathrm{i}}\mathrm{fro}{\mathrm{n}}_{\mathrm{i}}\in {\mathrm{N}}_{\mathrm{i}}\mathrm{and}{\mathrm{m}}_{\mathrm{j}}\in {\mathrm{N}}_{\mathrm{i}}. \\ \mathrm{Find}\mathrm{f}\left[\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)\left({\mathrm{m}}_1{\mathrm{m}}_2...\right)\right]\mathrm{as}: \\ \mathrm{f}\left[\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)\left({\mathrm{m}}_1{\mathrm{m}}_2...\right)\right]=\left({\mathrm{n}}_1{\mathrm{n}}_{2...}\right)\left({\mathrm{m}}_1{\mathrm{m}}_2...\right) \\ =\mathrm{f}\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)\mathrm{f}\left({\mathrm{m}}_1{\mathrm{m}}_2...\right) \\ \mathrm{Thus},\mathrm{f}\left[\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)\left({\mathrm{m}}_1{\mathrm{m}}_2...\right)\right]=\mathrm{f}\left({\mathrm{n}}_1{\mathrm{n}}_2...\right)\mathrm{f}\left({\mathrm{m}}_1{\mathrm{m}}_2...\right) \end{gathered}$$

Hence, is homomorphism and isomorphism.

Since is an isomorphism, it can be concluded that$\mathrm{G}\cong \sum _{\mathrm{ieI}}{\mathrm{G}}_{\mathrm{i}}$