Find a group $G$ and a subgroup $H$ such that $x,x',y,y'$ in $G$, $xH=x'H$ and $yH=y'H$, yet $xyH$ does not equal $x'y'H$ Find a group $G$ and a subgroup $H$ such that $x,x',y,y'$ in $G$, $xH=x'H$ and $yH=y'H$, yet $xyH$ does not equal $x'y'H$. (alternatively find an example of $G$ and $H$ where $(xH)(yH)$ does not equal $xyH$.
I haven't been able to find such examples. Can I get some help?
 A: Remember that $D_{4}$ can be generated by a rotating element $a$ and a fliping element $b$.  So, $D_{4}= \langle a,b | a^{4} = 1, b^{2}=1, ba = a^{-1}b \rangle $
Note that this also implies $ba^{-1} = ab$, $a^{-1}=a^{3} $, and $ ( a^{2} )^{-1} = a^{2} $.  
Let $ G = D_{4}$ and $ H = \{ 1, ba \} $.  $H$ is obviously a subgroup since $ba$ is an element of order 2.  
So, 
$aH= \{ a, aba \} = \{ a, ba^{-1} a \} = \{ a, b \} = \{ b, a \}  = \{ b, b^{2} a \} = bH$
Also, $ba^{2} H = \{ ba^{2}, ba^{2}ba \} = \{ ba^{2}, baba^{-1}a \} = \{ ba^{2}, bab \} = \{ ba^{2}, bba^{-1} \} = \{ ba^{2}, a^{-1} \}= \{ ba^{2}, a^{3} \}$.  
And also, $a^{3}H = \{ a^{3}, a^{3}ba \} = \{ a^{3}, a^{2} b a^{-1} a \}= \{ a^{3}, a^{2} b \}= \{ a^{3}, b a^{2} \}$.  So, $ba^{2} H = a^{3}H $.  
Let $x = a, x' = b, y = ba^{2}, y' = a^{3}$.
So, $aba^{2}H= b a^{-1} a^{2} H = baH =  \{ ba, baba \} = \{ ba, bba^{-1}a \} = \{ ba, 1 \} = H$.
And since $1 \in H \Rightarrow ba^{3} \in ba^{3}H$, however $ba^{3} \notin H $, then $abaH \neq ba^{3}H$ .  
Hence, $ xH = x'H$ and $y H = y'H $ but  $ xyH \neq x'y'H$.  Done!  
A: If $H\lhd G$, then always $xyH=xHyH=x'Hy'H=x'y'H$. If $G$ is abelian, then any subgroup is normal.
Thus we need a nonabelian group $G$ to start with. The smallest non-abelian group is $G=S_3$. It has (up to isomorphy and ignoring $1$ and $G$) two subgroups pf interest: $C_3=\langle (1\,2\,3)\rangle$ and $S_2=\langle (1\,2)\rangle$. But $C_3$ is normal. Therefore we try $H=S_2$.
What $x,x',y,y'$ should we pick?
To have $xH=x'H$ we must have either $x'=x$ or $x'=x(1\,2)$. In the frist case $yH=y'H$  implies $xyH=x'y'H$, so we rule that out. Apart from that, if $(x, x')$ works, than $(1,x^{-1}x')$ works as well. So we are stringly advised to take $x=1$, $x'=(1\,2)$.
It does not matter if $y'=y$ or not because $yH=y'H$ implies that $xyH\ne x'y'H$ also means $xyH\ne x'yH$. With our choices for $x,x'$ we thus look for $y$ with $yH\ne (1\,2)yH$. We should not pick $y\in H$ for that would make both sides $=H$. Why not try $y=(2\,3)$? Then $yH=\{(2\,3),(1\,3\,2)\}$ and $(1\,2)yH=\{(1\,2\,3),(1\,3)\}$, voila!
