Question

Let $\mathbf{I}$be the set of positive integers and assume that for each $\mathbf{i}$$\in$$\mathbf{I}$, is a group. The infinite direct product of the ${\mathbf{G}}_{\mathbf{i}}$is denoted $\underset{\mathbf{ieI}}{\mathbf{\prod }{\mathbf{G}}_{\mathbf{i}}}$and consists of all sequence $\left(a_1,a_2...\right)$with ${\mathbf{a}}_{\mathbf{i}}$$\in$${\mathbf{G}}_{\mathbf{i}}$. Prove that is a group under coordinate wise operation

$\left(a_{1,}{a}_2,...\right)\left(a_1,a_2,...\right)\mathbf{=}\left(a_1{b}_1,a_2{b}_2,...\right)\mathbf{.}$

.

Short answer

It has been proved that,$\underset{\mathrm{ieI}}{\prod {\mathrm{G}}_{\mathrm{i}}}$ is a group under coordinate wise operation.

Step-by-step solution

Given that

Given that $\mathrm{I}$is the set of positive integers and for each $\mathrm{i}\in \mathrm{I}$,${\mathrm{G}}_{\mathrm{i}}$ is a group.

The infinite direct product of the${\mathrm{G}}_{\mathrm{i}}$ is denoted$\underset{\mathrm{ieI}}{\prod {\mathrm{G}}_{\mathrm{i}}}$ and consists of all sequence $\left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right)$with ${\mathrm{a}}_{\mathrm{i}}\in {\mathrm{G}}_{\mathrm{i}}.$.

Prove the closure property

$$\begin{gathered} \left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right)and\left(b_1,b_2,...\right)arein\underset{ieI}{\prod G_i-,where{a}_i,b_i\in G_i.} \\ Then,\left(a_1,a_2,...\right)\left(b_1,b_2,...\right)=\left(a_1{b}_1,a_2{b}_2,...\right) \\ \sin ce{G}_{i}isagroup,a_i{b}_i\in G_i \\ so,\left(a_1{b}_1,a_2{b}_{2...}\right)\in \prod _{ieI}{G}_i. \\ Hence,\prod _{ieI}{G}_{i}satisfiestheclosureproperty. \end{gathered}$$

Prove the associativity

$$\begin{gathered} \mathrm{Consider}\left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right),\left({\mathrm{b}}_1,{\mathrm{b}}_2,...\right)\mathrm{and}\left({\mathrm{c}}_1,{\mathrm{c}}_2,...\right)\mathrm{are}\mathrm{in}\underset{\mathrm{ieI}}{\prod {\mathrm{G}}_{\mathrm{i}}}. \\ \mathrm{Prove}\mathrm{the}\mathrm{associativity}\mathrm{property}\mathrm{as}: \\ \left(\left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right)\left({\mathrm{b}}_1,{\mathrm{b}}_2,...\right)\right)\left({\mathrm{c}}_1,{\mathrm{c}}_2,...\right)=\left({\mathrm{a}}_1{\mathrm{b}}_1,{\mathrm{a}}_2{\mathrm{b}}_2,...\right)\left({\mathrm{c}}_1,{\mathrm{c}}_2,...\right) \\ =\left({\mathrm{a}}_1{\mathrm{b}}_1{\mathrm{c}}_1,{\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2,...\right) \\ =\left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right)\left({\mathrm{b}}_1{\mathrm{c}}_1,{\mathrm{b}}_2{\mathrm{c}}_2,...\right) \\ =\left({\mathrm{a}}_1,{\mathrm{a}}_2,...\right)\left(\left({\mathrm{b}}_1,{\mathrm{b}}_2,...\right)\left({\mathrm{c}}_1,{\mathrm{c}}_2,...\right)\right) \\ \mathrm{Thus},\prod _{\mathrm{ieI}}{\mathrm{G}}_{\mathrm{i}}\mathrm{satisfies}\mathrm{associative}\mathrm{property}. \end{gathered}$$

Find the identity and inverse element

It is clear that $\left(1,1,...\right)$, is the identity element and $\left(\frac{1}{{\mathrm{a}}_1},\frac{1}{{\mathrm{a}}_2},...\right)$is the inverse element. Thus, it can be concluded that $\underset{\mathrm{ieI}}{\prod {\mathrm{G}}_{\mathrm{i}}}$is a group under coordinate wise operation.