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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.

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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 $h'\in H$.

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  • $\begingroup$ Then we will have $h_1g = gh_2$ for some $h_1, h_2 \in H$, therefore $h_1 = gh_2g^{-1}$. Thank you, I got it now! $\endgroup$ – Mufei Li Sep 16 '16 at 17:29
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    $\begingroup$ @MufeiLi Not completely right. You have to start from $h\in H$ and $g\in G$ and prove that $ghg^{-1}\in H$. In the spoiler I do $g^{-1}hg\in H$, but of course it's the same. $\endgroup$ – egreg Sep 16 '16 at 17:32
  • $\begingroup$ I see, thanks for your patience and help! $\endgroup$ – Mufei Li Sep 16 '16 at 17:34

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