Question: Explain why (xH)g=(xg)H is not an action of a group G on the set of all left cosets of H in G
Im only just learning about acts on groups and need help understanding how to work this problem. So far what I understand is that in order for it to be an action of a group, it must satisfy the following:
1) must be well-defined
2) $ex=e$, for every $x \in S$
3) $(g_1g_2)x=g_1(g_2x)$
My Attempt
So, $G$ is a group and $H < G$ where $S={Hx/x\in G}$
$g(xH)=H(gx)$
To show that it $xH=x'H$ Then, $x^{-1}x' \in H$ Now, we must show $H(gx)=H(gx')$ for every $g \in G$ but $x^{-1}x' \in H$ so $gx'=(x^{-1}x')(gx) \in H(gx)$ So, $gx' \in H(xg)$ But certainly $H(gx')$ contains $e(gx')=gx'$. Thus, $H(gx)$ and $H(gx')$ have the elements $gx'$ in common. This proves that $H(gx)=H(gx')$
THis is where I have stopped.
A (left) group action on $L_H:=\{gH|g\in G\}$ is a function $a:G\times L_H\to L_H, (g,hH)\mapsto g\cdot hH(=ghH)$ such that the following two conditions are satisfied:
(i) $\forall g_1,g_2,h: a(g_1g_2,hH)=a(g_1,g_2\cdot hH)$
(ii) $\forall h:a(e,hH)=e\cdot hH=hH$.
For obvious reasons this action is called a left action, and it is the definition that you are citing. Similarly one can define a right action. So the issue would arise when commutativity is not satisfied by the given group (or in general when right cosets are not left cosets and vice versa (which happens when $H$ is not normal)). Also note that the map you give has domain $L_H\times G$.