For a finite group $G$ consider the left action of conjugation of $G$ on itself $·:G×G→G$ given by $g·x=gxg^{-1}$.

Question

For a finite group $G$ consider the left action of conjugation of $G$ on itself $·:G×G→G$ given by $g·x=gxg^{-1}$.

Prove that if $H$ is normal in $G$ then $H$ is the disjoint union of orbits of this action. (Note that the orbits in this case are called the conjugacy classes of G.)

Answer 1

Hint: take an element $h \in H$. What can you say about it's orbit?

More hints:

Note that by construction, orbits define an equivalence relation and thus $G$ is the disjoint union of the orbits of this action, $\mathcal{O}_1, \dots\mathcal{O}_n$. Now from these orbits we can select the ones that contain some element of $H$, call them $\mathcal{O}'_1, \dots, \mathcal{O}'_k$.

And finally,

We can prove that in effect, $H = \bigcup_{1 \leq j \leq k}\mathcal{O}'_j$ and the union will be disjoint as previously noted. Let's see both inclusions: take $h \in H$. Since the orbits of the action partition $G$, there exists some $s$ for which $h \in \mathcal{O}_s$. Since this orbit has an element of $H$, it necessarily equals some orbit $\mathcal{O}'_r$. Hence $h \in \bigcup_{1 \leq j \leq k}\mathcal{O}'_j$. Reciprocally, let $g \in \mathcal{O'}_j$ and (by construction) let $h \in \mathcal{O'}_j$ be the element of $H$ in this orbit. Since $g$ and $h$ belong to the same orbit we have that (for some $y$) $g = y \cdot h = y \cdot h \cdot y^{-1} \in yHy^{-1} = H$ as desired. Edit: as noted in the comments, we can quickly generalize this as follows; if we only require $H$ to be closed with respect to the action on $G$, this gives us the last inclusion (as $g = yh \in yH \subseteq H$). The rest works for any action and any set $H$. Hence any subset of $G$ closed with respect to an action on the group will be a (disjoint) union of orbits.

Answer 2

For any group action of the form $\boldsymbol{\cdot}$:$G$x$G \rightarrow G$, you may partition G into disjoint sets.

If you define $<x>=\{g\boldsymbol{\cdot}x: g \in G\}$, then you may then prove that distinct elements give rise to distinct orbits.

ie Suppose that $x,y \in G$, then $<x>=<y>$ or the orbits are disjoint.

It is sufficient to prove that if the orbits of x and y are not disjoint, then they are equal.

Suppose z is in both orbits. Then $z \in <x>$, so we may write $z=g \cdot x$, and $z \in <y>$, so we may write $z=h\cdot y$, for $h,g \in G$.

Then, $g \cdot x = h \cdot y$, so$ x=(g^{-1}h) \cdot y$, so $x \in <y>$.

Similarly, you may show $y \in <x>$, and therefore $<x>=<y>$.

If G is a finite group, then you may list the distinct orbits $<x_1>,...,<x_N>$, and it is quite clear the $G= \cup_{x \in G} <x>$, and since the orbits are disjoint, it follows that $G= \sqcup_{x \in G} <x>$.

We have now expressed G as a disjoint union of orbits of some representative elements in G. In order to write H, a normal subgroup, as a disjoint union, you simply need to choose the orbits that contain at least an element of H, and then you may write H as a disjoint union of those orbits.