Is a subgroup test necessary if proving normality?

Question

If I am proving normality, is it also necessary to first prove that the group is a subgroup? i.e. if $N$ is a subset of a group $G$ does $$ gNg^{-1} \subseteq N\, \forall\, g\in G \implies N\le G? $$

Answer 1

You must show that $N$ is a subgroup. $\{(123),(132)\}$ conjugates to itself by any element of $S_3$ and is not a subgroup.

Answer 2

I may be misunderstanding anything, but it seems to me that, if $N$ is a subset of $G$ which is not also a subgroup of $G$, then nothing can make it turn into a subgroup (let alone the normality condition), so your implication is false. On the other hand, if $N$ is a subset of $G$ which is also a subgroup of $G$, then your implication is vacuously true. So I guess that, perhaps, your real question was: if $N$ is a subgroup of $G$ such that, for some $H\le G$, the normality condition holds for every $h\in H$, does it follow that $N\le H$? The answer is no.

For a counter-example, take $G:=\operatorname{Sym}{(\mathcal G)}$ and $H:=\operatorname{Aut}{(\mathcal G)}$, for any group $\mathcal G$. Then, the isomorphic copy of $\mathcal G$ into $G$ by left multiplication ("Cayley's"), say $L$, fulfils the normality condition in $H$, but $L\not\subseteq H$.

Proof. For every $\sigma\in H$, $l_a\in L$, $g\in \mathcal G$: \begin{alignat}{1} (\sigma l_a\sigma^{-1})(g) &= \sigma(l_a(\sigma^{-1}(g))) \\ &= \sigma(a\sigma^{-1}(g)) \\ &= \sigma(a)g \\ &= l_{\sigma(a)}(g) \\ \end{alignat} whence: $\sigma l_a\sigma^{-1}\in L$, and finally: $\sigma L\sigma^{-1}\subseteq L$ for every $\sigma\in H$. But $L\cap H=\{\operatorname{id}_{\mathcal G}\}$; in fact: \begin{alignat}{1} l_a(gh)=l_a(g)l_a(h) &\iff agh=agah \\ &\iff a=e \\ &\iff l_a= \operatorname{id}_{\mathcal G} \\ \end{alignat} Therefore, despite $\sigma L\sigma^{-1}\subseteq L$ for every $\sigma\in H$, it is $L\not\le H$. $\space\Box$