Question

(a) Let G be a group and $\mathit{c}\mathbf{\in }\mathit{G}$. Prove that the map${\mathit{\theta }}_{\mathbf{c}}\mathbf{:}\mathbf{}\mathit{G}\mathbf{\to }\mathit{G}$ given by${\mathit{\theta }}_{\mathbf{c}}\left(x\right)\mathbf{=}\mathit{x}{\mathit{c}}^{\mathbf{-}\mathbf{1}}$ is an element of $\mathit{A}\left(G\right)$.

Short answer

It is proved that the function${\theta }_c:G\to G$ belongs to $A\left(G\right)$.

Step-by-step solution

Homomorphism Group

Suppose that $\left(G,*\right)$and $\left(H,◦\right)$are two groups. A function $\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{G}\mathbf{\to }\mathit{H}$is a homomorphism if role="math" localid="1653473858157" $\mathit{f}\left(a*b\right)\mathbf{=}\mathit{f}\left(a\right)\mathbf{◦}\mathbf{}\mathit{f}\left(b\right)$for all $\mathit{a}\mathbf{}\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$.

Prove that the map θc: G→G given by θc(x)=xc-1 is an element of A(G)

Consider a group G and an element $c\in G$.

Define a map${\theta }_c:G\to G$ by,

${\theta }_c\left(x\right)=x{c}^{-1}$

Assume the function${\theta }_c:G\to G$ is a bijection and hence, belongs to $A\left(G\right)$, the set of all permutations of G.

Consider$x,y\in G$ such that${\theta }_c\left(x\right)={\theta }_c\left(y\right)$ then, $x{c}^{-1}=y{c}^{-1}$.

Multiply by c in $x{c}^{-1}=y{c}^{-1}$.

$\begin{array}{rcl}x{c}^{-1}c& =& y{c}^{-1}c\\ x& =& y\end{array}$

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

Therefore, the function${\theta }_c:G\to G$ is one-one.

For any $x\in G$, the element$xc\in G$ such that,

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

Therefore, every element of G has a pre-image in G. Thus, the function${\theta }_c:G\to G$ is onto.

This proves that the function${\theta }_c:G\to G$ is one-one and onto and hence, bijection.

Therefore, it is proved that the function${\theta }_c:G\to G$ belongs to $A\left(G\right)$.