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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.

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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 $(1\ 3)$.

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  • $\begingroup$ +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.) $\endgroup$ Nov 17 '19 at 2:52
  • $\begingroup$ It seems you are referring to the last comment after the OP in the linked question. I think that argument is wrong, and I don't see how to salvage it. (Without doing something completely different.) $\endgroup$
    – verret
    Nov 17 '19 at 3:56
  • $\begingroup$ 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 $G\cong C_5\times A_4$. $\endgroup$
    – verret
    Nov 17 '19 at 3:59

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