Intuition behind coset equality

Question

We say $xH = yH$ iff $y^{-1}x \in H$. Writing out the definition of a coset, we have $xH := \{g\in G | g = xh, h \in H\}$, similarly $yH := \{g\in G | g = yh, h \in H\}$. If my logic is correct, $xH = yH$ implies $xh = yh$. So $y^{-1}xh = h$. I am not sure how to jump to $y^{-1}x \in H$ from this statement since we do not know if $H$ is abelian. What is the intuition behind this equality, and how can it be derived?

Answer 1

$xH=yH$ doesn't imply that $xh=yh$ necessarily. All you can deduce is that $xh_1=yh_2$ for $h_1, h_2 \in H$, ie it need not be the same element. then $xh_1=yh_2$ is equal to $y^{-1}x=h_2h_1^{-1} \in H$. Remember that $H$ is a subgroup of a group, so $h_1^{-1}$ exists, and by closure $h_2h_1^{-1}$ lies in $H$.

To show that $h$ need not be the same, look at $G=\mathbb{Z}/10\mathbb{Z}=\{0,1,2,3,4,5,6,7,8,9\}$, with $H=\{0,2,4,6,8\}$. Then $1H=\{0,2,4,6,8\}$, $2H=\{0,4,8,2,6\}$. Comparing elements, we have $1.H=2H$. For $h=4 \in H, x=1, y=2$, it doesn't hold that $1.4=2.4$, yet $xH=yH$.