I denote by $\leq$ subgroup. If $H\leq G$ and $g \in G$, how does it follow that if $gHg^{-1} \leq H$ then $H \leq g^{-1}Hg$?
I look first at an element $gh_ig^{-1}\in gHg^{-1}$, $h_i \in H$. We have that $gh_ig^{-1} = h_j$, for a unique $h_j \in H$ ($h_j$ is unique because $ghg^{-1} = gh'g^{-1}$ implies that $h=h'$), which gives $h_i = g^{-1}h_jg$. But then we get an equality, $H = g^{-1}Hg$, since every element $h_i \in H$ is also in $g^{-1}Hg$, and $|H| = |g^{-1}Hg|$.
What am I doing wrong?
Notice that $H = g^{-1}Hg$ is just a special case of $H \leq g^{-1}Hg$.
Nonetheless, you could argue as follows:
$$gHg^{-1} \leq H \implies gH \leq Hg \implies H \leq g^{-1}Hg$$