Question

Question: Let G be a subgroup of . Define a relation on the set$\left\{1,2,...,n\right\}$ by if$\mathit{a}\mathbf{=}\mathit{\sigma }\left(b\right)$ and $\mathit{a}\mathbf{~}\mathit{b}$only if for some$\sigma$ in G. Prove that $~$is an equivalence relation.

Short answer

Answer:

It is proved that given relation $~$is equivalence.

Step-by-step solution

Definitions of Equivalence relation

A relation R defined on a set is called an equivalence relation if it is reflexive, symmetric and transitive.

Prove that ~is an equivalence relation

Consider the symmetric group and suppose that be a subgroup of . Define a relation on the set$\left\{1,2,...,n\right\}$by $a~b$. If and only if $a=\sigma \left(b\right)$for some permutation a .

Objective is to prove that is an equivalence relation. That is to show that the relation is reflexive, symmetric and transitive.

First consider transitivity, note that there exists identity permutation $\sigma =I\in G$such that this implies that the relation is reflexive.

For Symmetric,

Let then there exist a permutation $\sigma \in G$such that $\sigma \left(b\right)=a$.Now since$\sigma \in G$ and G is a subgroup therefore inverse of $\sigma$exists and say it is$\alpha$ . Then $\sigma \alpha =\alpha \sigma =I$.

Now consider,

$\sigma \left(b\right)=a$

Multiply both sides by $\alpha$.

$\alpha \left(\sigma \left(b\right)\right)=\alpha \left(a\right)$

This gives,

$b=\alpha \left(a\right)$

This implies that$b~a$ .

Therefore, the relation $~$is symmetric.

For transitive,

Suppose that$a~b$ and $b~a$. Then there exist permutation and belongs to such that $a=\alpha \left(b\right)$and$b=\beta \left(c\right)$ .

Now consider,

$$\begin{gathered} a=\alpha \left(b\right) \\ =\alpha \left(\beta \left(c\right)\right) \\ =\alpha \circ \beta \left(c\right) \end{gathered}$$

Since, $\alpha$and$\beta$ belongs to , therefore, by closure axioms $\alpha \circ \beta \in G$. Thus, $a=\alpha \circ \beta \left(c\right)$, where $\alpha \circ \beta \in G$.

This shows that the relation$~$ is transitive as well.

Therefore, the given relation $~$is equivalence.