Question

Let $\mathbf{i}$be an integer with $\mathbf{1}\mathbf{\le }\mathit{i}\mathbf{\le }\mathit{n}$. Prove that the function ${\mathbf{\pi }}_{\mathbf{i}}\mathbf{:}{\mathbf{G}}_{\mathbf{1}}\mathbf{\times }{\mathbf{G}}_{\mathbf{2}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathbf{G}}_{\mathbf{n}}\mathbf{\to }{\mathbf{G}}_{\mathbf{i}}$

given by ${\mathbf{\pi }}_{\mathbf{i}}\left(a_1,a_2,a_3,...,a_n\right)\mathbf{=}{\mathbf{a}}_{\mathbf{i}\mathbf{}}$is a surjective homomorphism of groups.

Short answer

Answer:

It is proved that the function ${\mathrm{\pi }}_{\mathrm{i}}:{\mathrm{G}}_1\times {\mathrm{G}}_2\times ...\times {\mathrm{G}}_{\mathrm{n}}\to {\mathrm{G}}_{\mathrm{i}}$given by ${\mathrm{\pi }}_{\mathrm{i}}({\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,...,{\mathrm{a}}_{\mathrm{n}})={\mathrm{a}}_{\mathrm{i}}$is a surjective homomorphism of groups.

Step-by-step solution

Definition of Surjective Homomorphism

If a function is onto and homomorphism then, the function is a surjective homomorphism.

Prove that the function πi.G1×G2×...×Gn→G1 given by πi(a1,a2,a3,...,an)=ai is a surjective homomorphism of groups

Check for onto:

For any $a_i\in G$with $1\le i\le n$,there always exists $\left(a_1,a_2,...,a_n\right)\in G_1\times G_2\times ...\times G_n$such that ${\mathrm{\pi }}_{\mathrm{i}}\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)={\mathrm{a}}_{\mathrm{i}}$.

Therefore, ${\mathrm{\pi }}_{\mathrm{i}}:{\mathrm{G}}_1\times {\mathrm{G}}_2\times ...\times {\mathrm{G}}_{\mathrm{n}}\to {\mathrm{G}}_{\mathrm{i}}$is onto function.

Check for homomorphism${\mathrm{\pi }}_{\mathrm{i}}\left(\mathrm{\alpha }\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)+\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right)\right)=\mathrm{\alpha \pi }\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)+{\mathrm{\pi }}_{\mathrm{i}}\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right)$

Find ${\mathrm{\pi }}_{\mathrm{i}}\left(\mathrm{\alpha }\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)+\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right)\right)$as:

$$\begin{gathered} {\mathrm{\pi }}_{\mathrm{i}}\left(\mathrm{\alpha }\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)+\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right)\right)={\mathrm{\pi }}_{\mathrm{i}}(\left({\mathrm{\alpha a}}_1,{\mathrm{\alpha a}}_2,...,{\mathrm{\alpha a}}_{\mathrm{n}}\right)+\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right)) \\ ={\mathrm{\pi }}_{\mathrm{i}}\left({\mathrm{\alpha a}}_1+{\mathrm{b}}_1,{\mathrm{\alpha a}}_2+{\mathrm{b}}_2,...,{\mathrm{\alpha a}}_{\mathrm{n}}+{\mathrm{b}}_{\mathrm{n}}\right) \\ ={\mathrm{\alpha a}}_{\mathrm{i}}+{\mathrm{b}}_{\mathrm{i}} \\ ={\mathrm{\alpha \pi }}_{\mathrm{i}}\left({\mathrm{a}}_1,{\mathrm{a}}_2,...,{\mathrm{a}}_{\mathrm{n}}\right)+{\mathrm{\pi }}_{\mathrm{i}}\left({\mathrm{b}}_1,{\mathrm{b}}_2,...,{\mathrm{b}}_{\mathrm{n}}\right) \end{gathered}$$

Therefore, the function ${\mathrm{\pi }}_{\mathrm{i}}:{\mathrm{G}}_1\times {\mathrm{G}}_2\times ...\times {\mathrm{G}}_{\mathrm{n}}\to {\mathrm{G}}_{\mathrm{i}}$defined by ${\mathrm{\pi }}_{\mathrm{i}}({\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,...,{\mathrm{a}}_{\mathrm{n}})={\mathrm{a}}_{\mathrm{i}}$is a surjective homomorphism.