Equivalence relations, Cosets

Question

Let G be a group and for elements a,b (elements of)G let a R b mean that there exists an element x(element of)G such that a=xbx^(-1). Show that R is an equivalence relation on G.

Not really sure how to go about this one.

Answer 1

You need to check the three conditions for an equivalence relation:

1. Check that $x \sim x$ under your relation $R$. So is there an element $a$ in every group $G$ such that $x=a x a^{-1}$?

2. Now check that if $x \sim y$, that is if $x=aya^{-1}$ that $y=bxb^{-1}$ for some $b \in G$. For this try solving for $y$ and then relabeling to find your $b$'s in terms of your $a$'s. Then have you found elements that make this work?

3. Now check that if $x \sim y$, that is if $x=aya^{-1}$ and $y \sim z$, that is if $y=bzb^{-1}$, that $x \sim z$. That means that $x=czc^{-1}$ for some $c \in G$. Try solving for $z$ in terms of $b$ and $y$ and then substituting that in your equation for $x \sim y$. Once you have that equation, can you find an element $c$ in terms of $a,b$ so that you have $x \sim z$?