Why must a subgroup $H$ be a normal subgroup of $G$ so that the group operation $Hg Hf = H(gf)$ is well-defined?
In my notes I have written:
the operation must be well-defined so if $Hg_1 = Hg_2$ and $Hf_1 = Hf_2$ we must have $Hf_1g_1 = Hf_2g_2$.
But then I have written (which I do not understand):
So we have $g_2g_1^{-1} \in H$ and $f_2f_1^{-1} \in H$ (why is this) and we require $g_2f_2f_1^{-1}g_1^{-1}$ (why is this?).
If someone could explain this last line I'd be very grateful
Let's say the operation is well defined and let $g\in G,x\in H$. In this case we have:
$Hgxg^{-1}=HgHxHg^{-1}=HgHeHg^{-1}=Hgeg^{-1}=He=H$
Hence $gxg^{-1}\in H$ for all $g\in G,x\in H$. And this implies $H\trianglelefteq G$. So if $H$ is not normal in $G$ then there is no chance the operation will be well defined.