# Why require a normal subgroup to be also a subgroup, in the definition?

Let $$G$$ be a group, $$H\subset G$$ (but not necessarily $$H \leq G$$!) and, for all $$g\in G$$, $$gHg^{-1}\subset H$$. Is it true that $$H$$ is a subgroup of $$G$$? (And if it happens to be true, why to require a normal subgroup to be also a subgroup, in the definition?)

p.s. What follows is the contest from where this question originated (which can be skipped). This existential doubt came to me after reading a proof of the normality of the center $$Z(G)$$ of a group $$G$$. From the textbook:

Note that if $$z\in Z(G)$$, $$gzg^{-1}=z$$ for any $$g\in G$$. But then $$aZ(G)a^{-1}=Z(G)$$ given any $$a\in G$$. $$\square$$

As far as I can see, the author does not prove that $$Z(G)\leq G$$ (but this is evident, and maybe this is the reason why such a check is happily skipped). This made me wonder if there a more subtle reason to not explicitly write down it.

If $$G$$ is abelian, then $$gHg^{-1} = H$$ for all $$H \subset G$$ and $$g \in G$$. Thus, any subset $$H$$ that is not a subgroup provides a counterexample, when $$G$$ is abelian.
Let $$H$$ be the empty set. It is not a subgroup, but satisfies $$gHg^{-1}=H$$ (both are empty sets).
Edit: You want to make sure a "normal subset" (a subset $$H$$ satisfying $$gHg^{-1}=H$$) is a subgroup, because by the time you see quotient group, you will see that normal subgroups are the only objects over which you can take quotient of $$G$$.