Question

Let $\mathbf{H}$be a group and ${\mathit{T}}_{\mathbf{1}\mathbf{}}\mathbf{:}\mathbf{}\mathbf{H}\mathbf{}\mathbf{\to }{\mathbf{G}}_{\mathbf{1}}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{T}}_{\mathbf{2}}\mathbf{}\mathbf{:}\mathbf{}\mathbf{H}\mathbf{}\mathbf{\to }{\mathbf{G}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{T}}_{\mathbf{n}}\mathbf{:}\mathbf{}\mathbf{H}\mathbf{\to }{\mathbf{G}}_{\mathbf{n}}$, ,…, homomorphisms with this property: Whenever $\mathbf{G}$is a group and ${\mathbf{g}}_{\mathbf{1}}\mathbf{:}\mathbf{}\mathbf{G}\mathbf{\to }{\mathbf{G}}_{\mathbf{1}}\mathbf{,}{\mathbf{g}}_{\mathbf{2}}\mathbf{:}\mathbf{}\mathbf{G}\mathbf{\to }{\mathbf{G}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{g}}_{\mathbf{n}}\mathbf{}\mathbf{:}\mathbf{}\mathbf{G}\mathbf{\to }{\mathbf{G}}_{\mathbf{n}}$, ,…, are homomorphisms $\mathbf{g}\mathbf{}\mathbf{:}\mathbf{}\mathbf{G}\mathbf{}\mathbf{\to }\mathbf{H}$then there exists a unique homomorphism such that ${\mathbf{T}}_{\mathbf{i}}\mathbf{\circ }\mathbf{g}\mathbf{*}\mathbf{=}{\mathbf{g}}_{\mathbf{i}}$for every $\mathbf{i}$. Prove that $\mathbf{H}\mathbf{\cong }{\mathbf{G}}_{\mathbf{1}\mathbf{}}\mathbf{x}\mathbf{}{\mathbf{G}}_{\mathbf{2}}\mathbf{x}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{x}\mathbf{}{\mathbf{G}}_{\mathbf{n}}$. [See Exercise 19.]

Short answer

It has been proved that, $\mathrm{H}\cong {\mathrm{G}}_1\mathrm{x}{\mathrm{G}}_2\mathrm{x}...\mathrm{x}{\mathrm{G}}_{\mathrm{n}}$

Step-by-step solution

Given that

Given that $\mathrm{H}$is a group and ${\mathrm{T}}_1:\mathrm{H}\to {\mathrm{G}}_1,{\mathrm{T}}_2:\mathrm{H}\to {\mathrm{G}}_2,...,{\mathrm{T}}_{\mathrm{n}}:\mathrm{H}\to {\mathrm{G}}_{\mathrm{n}}$, ,…, homomorphisms with the following property:

Whenever $\mathrm{G}$is a group and ${\mathrm{g}}_1:\mathrm{G}\to {\mathrm{G}}_1,{\mathrm{g}}_2:\mathrm{G}\to {\mathrm{G}}_2,...,{\mathrm{g}}_{\mathrm{n}}:\mathrm{G}\to {\mathrm{G}}_{\mathrm{n}}$are homomorphisms, then there exists a unique homomorphism $\mathrm{g}*:\mathrm{G}\to \mathrm{H}$such that ${\mathrm{T}}_{\mathrm{i}}\circ \mathrm{g}*={\mathrm{g}}_{\mathrm{i}}$for every $\mathrm{i}$.

Prove homomorphism

Let $\mathrm{G}={\mathrm{G}}_1\mathrm{x}{\mathrm{G}}_2\mathrm{x}...\mathrm{x}{\mathrm{G}}_{\mathrm{n}}$with projections ${\mathrm{T}}_{\mathrm{i}}:\mathrm{H}\to {\mathrm{G}}_{\mathrm{i}}$

As per the property, there exists a unique $\mathrm{g}*:\mathrm{G}\to \mathrm{H}$with ${\mathrm{T}}_{\mathrm{i}}\circ \mathrm{g}*={\mathrm{g}}_{\mathrm{i}}$

Therefore, there exists a unique homomorphism $\mathrm{f}*:\mathrm{G}\to \mathrm{H}$with${\mathrm{g}}_{\mathrm{i}}\circ \mathrm{f}*={\mathrm{T}}_{\mathrm{i}}$

Prove isomorphism

Now $\mathrm{g}·\mathrm{f}·:\mathrm{H}\to \mathrm{H}\to \mathrm{H}$is a unique homomorphism with ${\mathrm{T}}_{\mathrm{i}}\circ \mathrm{g}*\mathrm{f}*={\mathrm{T}}_{\mathrm{i}}$.

This implies $\mathrm{g}*\mathrm{f}*=1_{\mathrm{H}}$

Similarly, $\mathrm{f}*\mathrm{g}*=1_{\mathrm{G}}$

Therefore, $\mathrm{f}*$and $\mathrm{g}*$are isomorphism

Thus, it can be concluded that$\mathrm{H}\cong {\mathrm{G}}_1\mathrm{x}{\mathrm{G}}_2\mathrm{x}...\mathrm{x}{\mathrm{G}}_{\mathrm{n}}$