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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?

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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 $ab\not\in Hxy$.

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\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 $G$).

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