# Prove the relation ~ on $A$ defined by $a$~$b$ if and only if $a = hb$ is an equivalence relation.

Let $H$ be a group acting of a set $A$. Prove that the relation ~ on $A$ defined by $a$~$b$ if and only if $a = hb$ for some $h \in H$ is an equivalence relation. (For each $x\in A$ the equivalence class of $x$ under class of $x$ under ~ is called the orbit of $x$ under the action of $H$. The orbits under the action of $H$ partition the set $A$.

proof: For the relation ~ to be an equivalence class, it must be reflexive, symmetric and transitive.

For reflexive, suppose for every $x \in A$ we have $x = 1* x$ so we have $x = x$ So $x$~$x$.

For symmetric, suppose for every $x,y \in A$ , if $x$ ~ $y$ we have $x = hy$ by the defined relation, where $h \in H$. Then $x = hy$ implies $h^{-1}x = h^{-1}(hy) = (h^{-1}h)y = 1*a = a$ by the definition of a group action. Thus $h^{-1}x = y$ implies $y$~ $x$.

for transitive: Suppose $x = hy$ and $y = kz$ . Then $x = hy = h(kz) = (hk)z$ So $x$~$y$ and $y~z$, we have $x$~$z$.

thus ~ is an equivalence relation.

Can someone please verify this is correct. If not can someone please give me some feedback. Thank you very much.

• This looks good to me. Commented Sep 10, 2015 at 1:17
• It's fine. It may be easier to write $ab^{-1} \in H$ than "$a=hb \text{ for some } h \in H$". Commented Sep 10, 2015 at 2:30

0) The neutral element on $H$ is the null/identity action on $A$. Proof: Let's assume it isn't. That is let's assume $1\in H$ acts on $a\in A$ as $1a=a'\neq a$, and that $g\in H$ is the identity transformation on $A$, i.e., $\forall a\in A\Longrightarrow ga=a$. Then applying $g$ on $a'$ we get $a'=ga'=g\,1\,a=g\,a=a$. But this is a contradiction, as we assumed $a'\neq a$. Hence, the neutral element in $H$, $1$ is indeed the identity action on $A$.
2) Symmetry: If $a\sim b\,\Longleftrightarrow\,\exists h\in H\,;\,a=hb\,\Longleftrightarrow\,h^{-1}a=b\,\Longleftrightarrow\,b\sim a$. QED. 3)Transitivity: (you got it perfect)