Let $H\le G$ then the following holds true:
$H$ is a normal subgroup of $G$ iff $\forall g,h\in G$ there is a $k\in G$ such that $gHhH=kH$.
Necessity is trival, but I'm stuck at the other way.
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1See this question, or many other ones here at this site for this question. May be, this one is a better duplicate? – Dietrich Burde May 03 '18 at 11:35
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I don't think it is the same. The existing $k$ need not to be $g$ as in your posted threads. – EpsilonDelta May 03 '18 at 12:18
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1Trivially $gh=gehe\in gHhH=kH$, so wlog $k=gh$. It's the same question. – ancient mathematician May 03 '18 at 13:37