Here's my problem...I have to ultimately show that $\varphi(hak)=\varphi(hbk)\Rightarrow hak=hbk$ and I'm having issues... I am told, both $H,K$ are subgroups of $G$ and $a\in G$. The goal is to show if $\varphi:HaK\rightarrow a^{-1}HaK, \varphi(hak)=a^{-1}hak$ is a bijective map. so I start with the one-to-one argument... $$\varphi(hak)=\varphi(hbk)$$ $$a^{-1}hak=b^{-1}hbk$$ $$hak=ab^{-1}hbk$$ And I'm stuck...what can i do to remedy this?
Also, is the surjective argument simply shown by the way the two sets are defined as $HaK$ and $a^{-1}HaK$? Or do we have to say let $a\in a^{-1}HaK$...then $$a=a^{-1}ah^{-1}hak^{-1}k$$ and deduce from that or something like that?
This doesn't depend on $H$ and $K$ being subgroups at all. If you take an arbitrary subset $X$ of $G$ and $a \in G$, then the map $X \to a^{-1} X$ defined by $x \mapsto a^{-1} x$ is a bijection. The inverse is, of course, given by $y \mapsto a y$.