Question

Prove that a $\mathit{h}\mathbf{}\mathbf{:}\mathbf{}\mathit{G}\mathbf{\to }\mathit{A}\left(G\right)$a given by $\mathit{h}\left(c\right)\mathbf{=}{\mathit{\theta }}_{\mathbf{c}}$, is an injective homomorphism of groups. Thus G is isomorphic to the subgroup Im h of A(G). This is the right regular representation of G.

Short answer

It is proved that$h:G\to A\left(G\right)$ given by$h\left(c\right)={\theta }_c$ is an injective homomorphism of groups.

Step-by-step solution

Homomorphism Group

Suppose that$\left(G,*\right)$ and$\left(H,◦\right)$ are two groups, then a function$\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{G}\mathbf{\to }\mathit{H}$ is a homomorphism if$\mathit{f}\left(a*b\right)\mathbf{=}\mathit{f}\left(a\right)\mathbf{◦}\mathbf{}\mathit{f}\left(b\right)$ for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$.

Prove that h : G→A(G) given by h(c)=θc is an injective homomorphism of groups

Define a function$h:G\to A\left(G\right)$ by,

$h\left(c\right)={\theta }_c$

Assume that the function$h:G\to A\left(G\right)$ is an injective group homomorphism.

To prove that$h\left(cd\right)=h\left(c\right)h\left(d\right)$ for all $c,d\in G$, prove that ${\theta }_{cd}={\theta }_c{\theta }_d$.

For fixed$c,d\in G$ and any $x\in G$,

$\begin{array}{rcl}{\theta }_{cd}\left(x\right)& =& x{\left(cd\right)}^{-1}\\ & =& x\left(d^{-1}{c}^{-1}\right)\\ & =& \left(x{d}^{-1}\right)c^{-1}\\ & =& {\theta }_c\left(x{d}^{-1}\right)\end{array}$

Simplify further,

$\begin{array}{rcl}{\theta }_{cd}\left(x\right)& =& {\theta }_c\left(x{d}^{-1}\right)\\ & =& {\theta }_c{\theta }_d\left(x\right)\\ {\theta }_{cd}& =& {\theta }_c{\theta }_{d\left(forallc,d\in G\right)}\end{array}$

Now consider,

$\begin{array}{rcl}h\left(cd\right)& =& {\theta }_{cd}\\ & =& {\theta }_c{\theta }_d\\ & =& h\left(c\right)h\left(d\right)\end{array}$

Therefore, the function$h:G\to A\left(G\right)$ is a homomorphism.

Consider$c,d\in G$ such that $h\left(c\right)=h\left(d\right)$. Then${\theta }_c\left(x\right)={\theta }_d\left(x\right)$ for all $x\in G$.

This implies$x{c}^{-1}=x{d}^{-1}$ for all $x\in G$.

For $x=e$,

$\begin{array}{rcl}e{c}^{-1}& =& e{d}^{-1}\\ c^{-1}& =& d^{-1}\\ c& =& d\end{array}$

Thus, no two elements of G are mapped to the same element of G.

Therefore, the function$h:G\to A\left(G\right)$ is an injective homomorphism, and by the first isomorphism Theorem,

$G\cong \mathrm{Im}h$

This proves the assumption.

Therefore, the required result is proved.