Question

Question: Prove Theorem 9.23.

Short answer

The theorem has been proved.

Step-by-step solution

Statement

Let $H$be a subgroup of a group $G$. Then, $H-$ conjugacy is an equivalence relation on the set of all subgroups of $G$.

If $A$ is $H$-conjugate to role="math" localid="1653323829998" $B-$ then, $A\square B$.

Prove reflexivity

Since $A=e^{-1}Ae-$then, $A\square A$.

Thus, reflexivity is done.

Prove symmetric

Let $A\square B$.

Then, $B=x^{-1}Ax$for some $x\in H$.

Further, $A=xB{x}^{-1}$and $B\square A$.

Thus, it is symmetric.

Prove transitivity

Let $A\square B$and $B\square C$.

Then, $B=x^{-1}Ax$and $C=y^{-1}By$for some $x,y\in H$.

Hence, ${\left(xy\right)}^{-1}A\left(xy\right)$and $xy\in H$.

Thus, $A\square C$.

Hence, transitivity is proved.

Therefore, $H-$ conjugacy is an equivalence relation on the set of all subgroups of $G$.