# Product of cosets if the subgroup is not normal

Let $$H$$ be a subgroup of a group $$G$$, $$H$$ is not necessarily normal.

Show that there exist $$a \in Hx$$ and $$b \in Hy$$ such that $$ab \notin Hxy$$ where $$x, y \in G$$ and $$Hx = \left \lbrace hx \vert h \in H \right \rbrace$$ are the right cosets.

$$ab \in (Hx) (Hy)$$ and if $$H$$ is normal, we have $$(Hx) (Hy) = Hxy$$.

In this case, since $$H$$ is not necessarily normal, I think that is the same that see that the product of two right cosets is not a right coset. And the answer it would be like here.

Is correct the reasoning?

Your reasoning is correct. In particular, the problem should really assume that $$H$$ is not normal. (Saying $$H$$ is "not necessarily normal" leaves open the possibility that it is normal, in which case there are no such $$a,b,x,y$$.)
If $$H$$ is not normal then there is some $$g\in G$$ such that $$gHg^{-1}$$ is not contained in $$H$$. In other words, $$gH$$ is not contained in $$Hg$$. So there is some $$h\in H$$ such that $$gh\not\in Hg$$. Now let $$x=g$$, $$y=e$$, $$a=g$$, and $$b=h$$. Then $$a\in Hx$$, $$b\in Hy$$, and $$ab\not\in Hxy$$.
\begin{alignat}{1} \forall x,y \in G, HxHy=Hxy &\Longrightarrow \forall x,y \in G, \forall h,h'\in H,\exists h''\in H\mid hxh'y=h''xy \\ &\Longrightarrow \forall x \in G, \forall h\in H,\exists h''\in H\mid hxh=h''x \\ &\Longrightarrow \forall x \in G, \forall h\in H,\exists h''\in H\mid xhx^{-1}=h^{-1}h'' \\ &\Longrightarrow \forall x \in G, \forall h\in H, xhx^{-1}\in H \\ &\Longrightarrow H\unlhd G \\ \end{alignat}
So, the normality of $$H$$ in $$G$$ is indeed necessary to get $$HxHy=Hxy$$ for every $$x,y\in G$$. Now, by taking the contrapositive:
\begin{alignat}{2} H\ntrianglelefteq G &\space\space\space\space\space\space\space\space\space\space\space\Longrightarrow &&\space\exists x,y\in G\mid HxHy\ne Hxy \\ &\stackrel{Hxy=Hx\{e\}y\space\subseteq\space HxHy}{\Longrightarrow} &&\space\exists x,y\in G, \exists h,h'\in H\mid hxh'y\ne h''xy, \forall h''\in H \\ &\space\space\space\space\space\space\stackrel{a:=hx, \space b:=h'y}\Longrightarrow &&\space\exists x,y\in G,\exists a\in Hx,\exists b\in Hy \mid ab\ne h''xy, \forall h''\in H \\ &\space\space\space\space\space\space\space\space\space\space\space\stackrel{}\Longrightarrow &&\space\exists x,y\in G,\exists a\in Hx,\exists b\in Hy \mid ab\notin Hxy \\ \end{alignat}
which is your claim (if you assume $$H$$ precisely not normal in $$G$$).