Question

Let $\mathit{G}$be a group and ${\mathit{f}}_{\mathbf{1}}\mathbf{:}\mathit{G}\mathbf{\to }{\mathit{G}}_{\mathbf{1}}\mathbf{,}{\mathit{f}}_{\mathbf{2}}\mathbf{:}\mathit{G}\mathbf{\to }{\mathit{G}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{f}}_{\mathbf{n}}\mathbf{:}\mathit{G}\mathbf{\to }{\mathit{G}}_{\mathbf{n}}$homomorphisms. For $\mathit{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{\dots }\mathbf{,}\mathit{n}$, let ${\mathit{\pi }}_{\mathbf{i}}$be the homomorphism of Exercise 8. Let ${\mathit{f}}^{\mathbf{\ast }}\mathbf{:}\mathit{G}\mathbf{\to }{\mathit{G}}_{\mathbf{1}}\mathbf{\times }\mathbf{\cdots }\mathbf{\times }{\mathit{G}}_{\mathbf{n}}$be the map defined by ${\mathit{f}}^{\mathbf{*}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{(}{\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}{\mathbf{a}}_{\mathbf{1}}\mathbf{\right)}\mathbf{,}{\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}{\mathbf{a}}_{\mathbf{2}}\mathbf{\right)}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathbf{f}}_{\mathbf{n}}\mathbf{\left(}{\mathbf{a}}_{\mathbf{n}}\mathbf{\right)}\mathbf{)}$.

Prove that ${\mathit{f}}^{\mathbf{\ast }}$is a homomorphism such that${\mathit{\pi }}_{\mathbf{i}}\mathbf{\circ }{\mathit{f}}^{\mathbf{\ast }}\mathbf{=}{\mathit{f}}_{\mathbf{i}}$ for each$\mathit{i}$ .

Short answer

It is proved that, $f^{\ast }$is a homomorphism such that${\pi }_i\circ f^{\ast }=f_i$ for each$i$ .

Step-by-step solution

Step-by-Step Solution Step 1: Define projection homomorphism

Definition of Group Homomorphism:

Let $(G,\ast )$and $(G\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$is said to be a group homomorphism if $f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

Let localid="1652780658503" $G$be a group and $f_i:G\to G_i$for $i=1,2,\dots ,n$ be the homomorphisms.

For $i=1,2,\dots ,n$, ${\pi }_i$ be the projection homomorphism ${\pi }_i:G_1\times G_2\times \cdots \times G_n\to G_i$defined by ${\pi }_i(a_1,a_2,\dots ,a_n)=a_i$.

Let$f^{\ast }:G\to G_1\times G_2\times \cdots \times G_n$ be the map defined by $f^{\ast }\left(a\right)=(f_1\left(a\right),f_2\left(a\right),\dots ,f_n\left(a\right))$.

Prove f∗ is a homomorphism

We have to prove that $f^{\ast }$is a homomorphism such that ${\pi }_i\circ f^{\ast }=f_i$, for each $i$.

Let $x,y\in G$.

Find $f^{\ast }\left(xy\right)$as:

$$\begin{gathered} f^{\ast }\left(xy\right)=(f_1\left(xy\right),f_2\left(xy\right),\dots ,f_n\left(xy\right)) \\ =(f_1\left(x\right)f_1\left(y\right),f_2\left(x\right)f_2\left(y\right),\dots ,f_n\left(x\right)f_n\left(y\right)) \\ =(f_1\left(x\right),f_2\left(x\right),\dots ,f_n\left(x\right))(f_1\left(y\right),f_2\left(y\right),\dots ,f_n\left(y\right)) \\ =f^{\ast }\left(x\right)f^{\ast }\left(y\right) \end{gathered}$$

Therefore, $f^{\ast }\left(xy\right)=f^{\ast }\left(x\right)f^{\ast }\left(y\right)$

Hence,$f^{\ast }$ is a homomorphism.

Prove πi∘f∗=fi for each  i

Let$x\in G$ .

Find ${\pi }_i\left(f^{\ast }\left(x\right)\right)$as:

$\begin{array}{c}{\pi }_i\left(f^{\ast }\left(x\right)\right)={\pi }_i\left((f_1\left(x\right),f_2\left(x\right),\dots ,f_n\left(x\right))\right)\\ =f_i\left(x\right)\end{array}$

Thus, ${\pi }_i\circ f^{\ast }=f_i$, for each$i$ .

Hence,$f^{\ast }$ is a homomorphism such that${\pi }_i\circ f^{\ast }=f_i$ for each$i$ .