I am learning group theory and trying to practice the concepts I am learning. In particular, I learned that $H$ is a normal subgroup of $G$ if and only if $ghg^{-1}\in H$ for all $g\in G$ and for a given $h\in H$, but I also learned the equivalent following definition: $H$ is a normal subgroup of $G$ if and only if $H$ is a union of conjugacy classes of $G$. I think I'm having trouble grasping this rephrasing because when I try to apply it, I get the wrong results.
For example, given the group $S_3$, I want to find its normal subgroups, thus I determined that $S_3$ has three conjugacy classes, namely $\{(12), (13), (23)\}$, $\{(123), (132)\}$ and of course $\{()\}$ (I will write the identity element as the empty permutation).
I then used the second definition and got the following sets:
- $\{()\}$
- $\{(12), (13), (23)\}$
- $\{(123), (132)\}$
- $\{(), (12), (13), (23)\}$
- $\{(), (123), (132)\}$
- $\{(), (12), (13), (23), (123), (132)\}$
- $\{(12), (13), (23), (123), (132)\}$
Obviously, few of them are normal $S_3$ subgroups! Typically the sets that do not have the identity $()$ are not groups at all. I know from many ressources that $S_3$ has only three normal subgroups, the trivial group, itself and $\{(), (123), (132)\}$.
I can probably imagine that it is implied that the union must be made with the class of the identity, so as to at least have groups, but even if I do that I still have a group that is not invariant to $S_3$, namely $\{(), (12), (23), (13)\}$.
What is wrong with my understanding of this definition?
Neither of your characterisations of normal subgroup are correct. You should re-read whatever sources you are using carefully.
A subgroup $H$ of $G$ is normal if and only if for all $h \in H$, for all $g \in G$, we have $ghg^{-1} \in H$ (compared to what you wrote, it's not just "a given" $h\in H$, but all $h\in H$).
If $H \leqslant G$, then $H$ is normal if and only if it can be written as a union of conjugacy classes. Unions of conjugacy classes, even ones containing the identity, aren't in general subgroups (as your example shows).