# Left coset as group action on a subgroup

Take a group $$G$$ and a subgroup $$H$$. Let's define the set of left cosets of H as: $$S=\{aH | a\in G\}$$ It's pretty clear to me we can define a left action from G to S like this:$$g(aH)=gaH$$ This basically permutes the cosets while G is acting on them. However, I need a confirmation with a simpler question (probably trivial for this space). The way in which we defined the left cosets set, actually suggests we are letting elements of $$G$$ running on the subgroup $$H$$, obtaining all cosets.

I mean, it sounds a bit if we could identify the cosets as orbits of this "action". The wrong part affecting this "feeling" seems to me this cannot be called an action at all, technically.

An action is an homomorphism from G to Perm(S), where S is the "target" set the group acts onto.

So here we are somewhat letting G acting on H producing new cosets, but the result is not any permutation of H. So this is not a homomorphism.

So the only left action I can see related to this is the one already considering the whole set of left cosets as target.

Am I correct in solving my doubts?

• I don't really understand what you're looking for, but if $H$ acts on $G$ on the right by $G\times H\to G, (g,h)\mapsto gh$, then the orbits of this right action are precisely the cosets $gH$. Now if you don't like right actions, you can make this a left action by $H\times G\to G, (h,g)\mapsto gh^{-1}$, it doesn't change much. So $G/H$ is the set of orbits of $G$ under the action of $H$ – Max Jan 29 at 11:18
I don't really understand what you're looking for, but if $$H$$ acts on $$G$$ on the right by $$G\times H\to G,(g,h)\mapsto gh$$, then the orbits of this right action are precisely the cosets $$gH$$. Now if you don't like right actions, you can make this a left action by $$H\times G\to G,(h,g)\mapsto gh^{−1}$$, it doesn't change much. So $$G/H$$ is the set of orbits of $$G$$ under the action of $$H$$