Suppose $H$ is a subgroup of $G$ and that $K$ is the set of left cosets $xH$ of $H$ in $G$. I'm asked to show that $g\cdot xH:=gxH$ is an action of $K$ on $G$ and that it is well defined.
How do I show that $1_G\cdot xH=xH$ and that $gh\cdot xH=g\cdot (hxH)$ for all $xH\in K$? Is this all I need to show? It seems obvious that these are true but not sure what reasoning to give. How do I show it's well defined?
Can I just say $1_G\cdot xH=1_GxH=xH$ and $gh\cdot xH=ghxH=g\cdot (hxH)$?
On
You are using a specific representation of a coset ("$xH$") to define a map whose argument is a pair $(an \space element \space of \space G, \space a \space \space coset \space \space as \space such)$. We want this arbitrary choice not to interfere with the result of the mapping ("good definiteness"), namely we want that $$y \in xH \Rightarrow g \cdot yH = g \cdot xH$$
Suppose $xH=yH$, that is, $x^{-1}y\in H$. To verify that the action is well-defined, we need to show that $gxH=gyH$. It’s easy to see that $$(gx)^{-1}gy=x^{-1}y\in H.$$ Thus the action is well-defined.