Let's consider two subsets of the symmetric group $S_9$, $H=\{(123)^a(456)^b(789)^c: 0\le a,b,c \le 2 \}$ and $K=\{(147)^d(258)^d(369)^d:d=0,1,2\}$. In fact $H$ and $K$ are subgroups of $S_9$. Furthermore, $HK=\{hk:h\in H,k\in K\}$ is also a subgroup of $S_9$ with an order $81$. Is $K$ normal in $HK$? Is $HK$ normal in $S_9$?
I know that $H$ is normal in $HK$ and $K$ is not normal in $S_9$.
And I think $K$ is not normal in $HK$ by the following example $$ (123)(147)(258)(369)(123)^{-1}=(169)(247)(358)$$ However, I am struggling to show the element on the RHS is not in $HK$
To show $HK$ is not normal in $S_9$, I am using the following example $$(15)(123)(15)^{-1}=(235)$$ Again I don't know how to show the RHS element is not in $HK$. My thought is to take any element $hk$. The result of this product is always a cycle of 9 so elements generated by $hk$ is always either 3 distinct cycles of 3 or a cycle of 9. So RHS element is not possible.
You've stated that you know that $H$ is normal in $HK$.
Noting that $H,K$ are disjoint (i.e., $H\cap K=\{e\}$), if $K$ was normal in $HK$, it would follow that $hk=kh$ for all $h\in H,\,k\in K$.
Using the properites of normal subgroups to show that two elements commute
But if we take $h=(123)$ and $k=(147)(258)(369)$, it's easily verified that $hk\ne kh$.
Hence $K$ is not normal in $HK$.
Next consider the question of whether $HK$ is normal in $S_9$.
Since $3$-cycles are even permutations, it follows that all elements of $HK$ are even permutations.
Hence $HK$ is a subgroup of $A_9$.
But then if $HK$ was normal in $S_9$, it would also be normal in $A_9$, contrary to the fact that $A_9$ is simple.
Hence $HK$ is not normal in $S_9$.