Let $G$ be a group and $H \unlhd G$ and $K_1,K_2 \leq G$. Assume $K_1 \neq K_2$ and $H \cap K_1=H \cap K_2=\{1\}$.
My question is, are the above assumptions enough to guarantee $HK_1 \neq HK_2$? If not, what additional assumptions would be needed?
My motivation for asking this question is Classifying groups of order 60. The OP justified one of the steps in his/her work in a comment, and I am trying to follow the logic. Perhaps I would need to add the assumption that $H$ is cyclic? Or that $K_1$ and $K_2$ are? Or that $|K_1|=|K_2|$? Or that $H \leq Z(G)$?
I have tried making the assumptions at the top of the page, assuming in addtion that $HK_1=HK_2$ and trying to find a contradiction. I see that $HK_1=HK_2$ ensures $HK_1 \cap HK_2=HK_1$, which seems fishy since $K_1 \neq K_2$, but I don't know where to go next.
Let $G$ be the group of the square. Let $H$ be the group generated by rotation halfway round. Let $K_1,K_2$ be generated by the reflections in the two diagonals. Then $H$ is normal in $G$, $K_1\ne K_2$, $H\cap K_1=H\cap K_2=\{\,1\,\}$, but $HK_1=HK_2$ is the group generated by both diagonals.
Another example. $G=S_3$, $H=A_3$, $K_1$ generated by $(1\ 2)$, $K_2$ generated by $(1\ 3)$.