# Do these assumptions guarantee $HK_1 \neq HK_2$?

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 $$(1\ 3)$$.
• +1 So I take it the answer is "no." Might I ask, what extra assumptions would make the answer "yes?" (The particular case floating in the back of my head is $|G|=60$, $|H|=5$, $|K_1|=|K_2|=4$. And $G$ doesn't have a normal Sylow-3 subgroup.) Nov 17 '19 at 2:52
• Here's an alternate argument for that point in the proof. We have $G/P\cong A_4$. Now, let $C$ be the centraliser of $P$ in $G$. This is a normal subgroup of $G$ containing $P$. By the $N/C$ theorem, $G/C$ embeds in $Aut(P)=Aut(C_5)\cong C_4$. Since $G/P\cong A_4$, the only possibility is that $G/C=1$ so $P$ is central in $G$. It's then not hard to finish from there and conclude that $G\cong C_5\times A_4$. Nov 17 '19 at 3:59