Let $H$ be a subgroup of the Group $G$. Then $H$ is normal in $G$ if and only if it can be expressed as a union of conjugacy classes.

Proof

$H$ is normal in $G$ if and only if $gHg^{-1} = H$ for all $g \in G$; equivalently, if and only if $ghg^{-1} \in H$ for all $h \in H$ and $g \in G$.

But if we fix $h \in H$, then the statement that $ghg^{-1} \in H$ for all $g \in G$ is equivalent to insisting that the conjugacy class of $h$ is contained in $H$. Therefore $H$ is normal in $G$ if and only if, for all $h \in H$, the conjugacy class of $h$ lies in $H$.

If $H$ is normal, then it is clearly a union of conjugacy classes (namely $\cup_{h \in H} C_h$, where $C_h$ is the conjugacy class of $h$).

Conversely, if $H$ is not normal, then there is some $h \in H$ such that the conjugacy class of $h$ is not wholly in $H$; so $H$ is not a union of conjugacy classes because it contains $h$ but not the entire conjugacy class of $h$. (Here we have used that the [-conjugacy_classes_partition_the_group].)

Interpretation

A normal subgroup is one which is fixed under conjugation; the most natural (and, indeed, the smallest) objects which are fixed under conjugation are conjugacy classes; so this criterion tells us that to obtain a subgroup which is fixed under conjugation, it is necessary and sufficient to assemble these objects (the conjugacy classes), which are themselves the smallest objects which are fixed under conjugation, into a group.