Question

Let G be a group and define $\mathit{a}\mathbf{~}\mathit{b}\mathbf{}$if and only if there exists $\mathit{c}\mathbf{\in }\mathit{G}$such that $\mathit{b}\mathbf{=}{\mathit{c}}^{\mathbf{1}}\mathit{a}\mathit{c}$. Prove that$\mathbf{~}$is an equivalence relation on G.

Short answer

It is showed that $~$is the equivalent relation on G.

Step-by-step solution

Write the given data from the question.

The Gis the group and defines$a~b$

The expression,$b=c^{-1}ac$,$c\in G$

Step 2:  Prove that ~ is an equivalence relation on G

To prove that$~$is an equivalent reaction, the following should satisfy.

(i) localid="1659178138349" $\mathbf{~}$is reflexive

(ii) localid="1659178133377" $\mathbf{~}$is symmetric

(iii)$\mathbf{~}$is transitive

(i) To prove $~$is reflexive.

Let assume $a\in G$then$a~a$, thus $a=e^{-1}ae$

From the above G is a group and exist.

Therefore,$~$is reflexive.

(ii) To prove $~$is symmetric.

Let $a,b\in G$and $a~b$, there exist$c\in G$such that

Thus$b=c^{1}ac$

Multiply byc both the sides.

$cb=c{c}^{-1}ac$

Multiply by $c^{-1}$both the sides.

$$\begin{gathered} cb{c}^{-1}=c{c}^{-1}ac{c}^{-1} \\ cb{c}^{-1}=eae \\ cb{c}^{-1}=e \end{gathered}$$

Thus,$a=c{\mathrm{bc}}^1$, where $c\in G$.

Therefore,$~$is symmetric.

(iii) To prove $~$is transitive.

Let $a,b,c\in G$$a,b,c\in \mathrm{c}G$and $a~b,b~c$

So,$c,d\in G$such that,$b=c^{1}ac$and $c=d^{1}b\mathrm{d}$.

For $g\in G$,$c=g^{1}ag$

$$\begin{gathered} c=d^{-1}\left(c^{-1}ac\right)d \\ c=d^{-1}{c}^{-1}acd \\ c=\left(cd{)}^{-1}a\right(cd) \end{gathered}$$

Substitute g for cd into above equation.

$c=g^{1}ag$

Since $c,d\in G$ and g is group,$cd\in G$and also ${\left(cd\right)}^{-1}$exist.

Therefore, $~$is transitive.

Hence the three conditions are satisfied therefore,$~$is equivalent relation on G.