Prove that the double cosets $HgH$ are the left cosets $gH$ iff $H$ is normal

Question

Definition: Let $H$ and $K$ be subgroups of a group $G$, and let $g$ be an element of $G$. The set $HgK = \{x\in G \mid x = hgk, h\in H, k\in K\}$ is called a double coset.

Theorem: Double cosets partition $G$.

Question: Let $H$ be a subgroup of a group $G$. Show that the double cosets $HgH = \{ h_1 g h_2\mid h_1, h_2 \in H\}$ are the left cosets $gH$ iff $H$ is normal.

Attempt:For the direction from the left to the right, it is easy to prove as $h_1gh_2$ = $gh_3$, for some $h_1,h_2,h_3 \in H$, $g\in G$, then $h_1=gh3h_2^{-1}g^{-1}$, so $H$ is normal. But for the converse direction, I'm stuck then. I know I have to prove that $HgH \subset gH$ and $gH \subset HgH$ so $HgH = gH$, with the fact that $H$ is normal, but I don't know how can I get started.

Answer 1

Note that $gH\subseteq HgH$ and $Hg\subseteq HgH$. Next, note that if $H$ is normal, we have $$ h_1gh_2= g\underbrace{(g^{-1}h_1g)}_{\in H}h_2 $$ and therefore $HgH\subseteq gH$.

Suppose conversely that $HgH=gH$, for every $g$. Since $Hg\subseteq HgH$, we have $Hg\subseteq gH$. Can you end without looking at the spoiler?

Then, if $h\in H$ and $g\in G$, we have $g^{-1}hg=g^{-1}gh'=h'$ for some $h'\in H$.