Question

Question: Let G be a group and let A(G) be the group of permutations of the set G. Define a functionfrom G to A(G) by assigning to each $d\in G$the inner automorphism induced by (as in Example 9 with c=d^-1 ). Prove that g is a homomorphism of groups.

Short answer

Answer

It is proved that g is a homomorphism of groups

Step-by-step solution

Homomorphism Groups

$$\begin{gathered} \mathit{S}\mathit{u}\mathit{p}\mathit{p}\mathit{o}\mathit{s}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathbf{(}\mathit{G}\mathbf{,}\mathbf{*}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{(}\mathit{H}\mathbf{,}\mathbf{\circ }\mathbf{)}\mathbf{}\mathbf{}\mathit{a}\mathit{r}\mathit{e}\mathbf{}\mathit{t}\mathit{w}\mathit{o}\mathbf{}\mathit{g}\mathit{r}\mathit{o}\mathit{u}\mathit{p}\mathit{s}\mathbf{,}\mathbf{} \\ \mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathit{a}\mathbf{}\mathit{f}\mathit{u}\mathit{n}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}\mathbf{}\mathbf{}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{h}\mathit{o}\mathit{m}\mathit{o}\mathit{m}\mathit{o}\mathit{r}\mathit{p}\mathit{h}\mathit{i}\mathit{s}\mathit{m}\mathbf{} \\ \mathbf{\text{iff(a*b)=f(a)∘f(b),foralla,b∈G}} \end{gathered}$$

Prove that g is a homomorphism of groups.

Consider a group G and group A (G) of permutations of the set G.

Define a function $g:G\to A\left(G\right)$such that every element localid="1651730937297" $d\in G$ is mapped to the inner automorphism induced by d^-1. That is,

localid="1659354878466" $$\begin{gathered} {\text{g(d)=f}}_d \\ Here,f_d=G\to Gisafunctiondefinedby{f}_d\left(x\right)=dx{d}^{-1}. \\ {\text{Assumethatthefunctiong:G→A(G)definedbyg(d)=f}}_{d}isahomomorphism. \\ Toprovethis,showthat{f}_{ab}=f_a{f}_{b}forallab\in G. \\ Now,foranyx\in G \\ {f}_{ab}\left(x\right)=\left(ab\right)x{\left(ab\right)}^{-1} \\ =a\left(bx{b}^{-1}\right)a^{-1} \\ =f_{a}bx{b}^{-1} \\ =f_a{f}_b\left(x\right) \end{gathered}$$

Therefore, for all $f_{ab}\left(x\right)=f_a{f}_b\left(x\right)$. This implies that$f_{ab}=f_a{f}_b$.

Now consider,

$$\begin{gathered} g\left(xy\right)=f_{xy} \\ =f_x{f}_y \\ =g\left(x\right)g\left(y\right) \\ \text{Therefore,g(xy)=g(x)g(y)forallx,y∈G.} \\ \text{Thus,thefunctionishomomorphism.Thisprovestheassumption}. \end{gathered}$$

Hence, it is proved that is a homomorphism of groups.