# Conjugacy classes

I don't seem to be able to follow this part of the proof. Why are we able to say $x \sim y$? Why does it suffice to prove that $y \in H$?

Let $G$ be group.

$(a)$ We say that elements $x,y \in G$ are conjugate (or more precisely are conjugate in $G$) if there exists $g \in G$ with $g^{-1}xg=y$. Prove that conjugacy is an equivalence relation.

$(b)$ There equivalence classes in $(a)$ are called conjugacy classes. Prove that a subgroup $H$ of $G$ is normal iff it is a union of conjugacy classes.

$(b)$ Assume first that $H$ is normal. Say $x \sim y$ and $x \in H$. It suffices to prove that $y \in H$. By definition of $\sim$, there exists $g$ such that $y=g^{-1}xg$. Hence $y\in g^{-1}Hg=H$ by normality, and we're done.

(Original pictures here and here)

You have to show that normal subgroups contain entire conjugacy classes (perhaps one, perhaps a union of several). So assuming $x\in H$ is an element of the subgroup and assuming $y\sim x$ is in $x$'s conjugacy class, we need to make sure that $y\in H$. Then $H$ must contain all elements which are "equivalent" to each other (meaning they are conjugates of each other).
You are able to assume $y \sim x$ at this point of the proof because "$y \sim x$" is equivalent to the statement "$y$ is an element of the conjugacy class of $x$". Let me write this all out carefully.
Assuming $H$ is a normal subgroup of $G$, you must prove $H$ is a union of conjugacy classes. To prove that $H$ is a union of conjugacy classes, the strategy will be to prove that if $x \in H$ then the conjugacy class of $x$ is a subset of $H$; once that is done, then you will know that $H$ is the union of the conjugacy classes each of the elements of $H$, and the proof is finished.
So, assuming that $x \in H$ and assuming that $y$ is in the conjugacy class of $x$, you must prove that $y \in H$. Translation: assuming that $x \in H$ and assuming that $y \sim x$, you must prove that $y \in H$.