Why is H normal in G equivalent to the conjugate of H by g being a subset of H?

Question

Let $H\le G$

Why are the following equivalent?

i) $H \unlhd G $

ii) $ gHg^{-1} \subseteq H$ for all $g \in G$

In other words, why do we not need to show that $H \subseteq gHg^{-1}$ for all $g \in G$ ?

Answer 1

The trick here is that ii) holds for all $g \in G$, and you must use the containment at two different elements of $G$. More precisely, fix a $g \in G$ and assume ii) that $xHx^{-1} \subseteq H$ for all $x \in G$. Setting $x = g^{-1}$ yields the containment

$$ xHx^{-1} = g^{-1}Hg \subseteq H, $$

and multiplying the left and right of this containment by $g$ and $g^{-1}$, respectively, we conclude $H \subseteq gHg^{-1}$.