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We can remark that the left coset $gH$ of $g \in G$ relative to a subgroup $H$ of $G$ is the orbit of $g$ under the action of $H \subset G$ acting by right translation.

What is that right translation? and how can I prove that the orbit of $g$ under the action of $H \subset G$ acting by right translation is $gH$ ?

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    $\begingroup$ Right translation just means multiplying from the right by an element of $H.$ So you just start with $g,$ and see what you can get by hitting it from the right with elements of $H.$ $\endgroup$ Commented Mar 30, 2013 at 23:14
  • $\begingroup$ Translation, as a term, is more useful in abelian groups where the coset $gH$ would be denoted by $g+H$. $\endgroup$ Commented Mar 30, 2013 at 23:56

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Right translation can equally be read as "right multiplication", except there is an implication of commutativity.

As to your second query, let the subgroup $H$ act on $G$ by right multiplication: $$h \cdot g = gh \qquad \forall g \in G \quad \forall h \in H$$ For any $g \in G$, the orbit $H \cdot g$ is the set $$\{ h \cdot g : h \in H \} = \{ gh: h \in H \} = gH.$$ Note that we didn't need the operation to be commutative here.

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