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.
Best Answer
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.