Normal Subgroup Test

Theorem: If $$G$$ is a group and $$H \le G,$$ then $$H$$ is a normal subgroup of $$G$$ if and only if $$xHx^{-1} \subseteq H$$ for all $$x \in G$$.

Proof
Suppose $$H$$ is normal in $$G$$. Then for each $$x \in G$$ and $$a \in H,$$ $$\exists b \in H$$ such that $$xa=bx$$. Thus, $$xax^{-1}=b \in H$$ implying $$xHx^{-1} \subseteq H$$.

Conversely, suppose $$xHx^{-1} \subseteq H$$ for all $$x \in G$$. Then for $$g \in G,$$ we have $$gHg^{-1} \subseteq H,$$ which implies $$gH \subseteq Hg$$. Also, for $$g^{-1} \in G,$$ we have $$g^{-1}H(g^{-1})^{-1}=g^{-1}Hg \subseteq H$$ which implies $$Hg \subseteq gH$$. Therefore, $$gH=Hg$$ meaning $$H \triangleleft G$$.