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.
Your proof of symmetry is confusing, at best. The rest is good.
Let's keep the details under control:
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 $A$.
1) Reflexivity: (you got it perfect)
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)