Is this action well defined?

Let $ H $ be a proper subgroup of $ G $ . Then is \begin{align*} \rho :G\times (G:H)& \rightarrow (G:H) \\ (g,hH)& \mapsto ghH \end{align*} a well defined action? This is the left multiplication action and it seems to satisfy all conditions except for the fact that I am not sure that this is even a proper map. Let me elaborate. As $ H $ is not necesseraly normal, coset multiplication is not well defined. Thus if we consider \begin{align*} h_1 H=h_2 H, \end{align*} i.e. $ h_1 h_2 ^{-1}\in H $, we need not have \begin{align*} \rho (g,h_1 H)=\rho (g,h_2 H) \tag{\(*\)} \end{align*} as \begin{align*} (*)& \iff gh_1 H=gh_2 H \\ & \iff gh_1 ( gh_2 )^{-1}\in H \\ & \iff gh_1 h_2 ^{-1} g^{-1} \in H \end{align*} need not be true as $ H $ is not neceserally a normal subgroup.

Main Question: Is the action well defined?

I would really appreciate if someone can explain this to me.

Solution 1:

You messed up the condition for equality of left cosets.

Recall that $xH=yH$ if and only if $y^{-1}xH=H$, if and only if $y^{-1}x\in H$.

So if $xH=yH$ and $g\in G$, then $y^{-1}x\in H$, hence $y^{-1}(g^{-1}g)x\in H$, hence $(y^{-1}g^{-1})(gx)\in H$, hence $(gy)^{-1}(gx)\in H$, hence $gxH = gyH$. So the action is well defined.