If $xH=yH$, then $xy^{-1} \in H$. Use a counterexample to disprove the following statements: 


*

*If $xH=yH$, then $xy^{-1} \in H$.

*If $xy^{-1} \in H$, then $xH=yH$.


I was thinking for the first statement:
$$xH=yH \rightarrow y^{-1}x \in H$$
But we do know if H is commutative. Is this correct?
Would it be the same for the second statement.
 A: Consider the Dihedral group $D_3 = \langle r,s\,|\,r^3=s^2=1, rs=sr^2\rangle$ of order six and its subgroups $H = \{ 1,s\}$ and $H' = \{1,sr\}$. Let $x = r$ and $y=sr^2$. Then
(1) $xH = \{ r,sr^2\} = yH$ but $xy^{-1} = sr \notin H$;
(2) $xy^{-1} = sr\in H'$ but $xH' = \{r,s\} \neq \{r^2,sr^2\} = yH'$.
A: I don't know any group theory. Here's how I figured out a counterexample to your first statement, just knowing the basic definitions, and following my nose.
First, I figured out that $xH=yH$ means $y^{-1}x\in H$. So I'm looking for an example where $y^{-1}x\in H$ while $xy^{-1}\notin H$. So $x$ and $y$ should not commute.
Trying to keep things simple, I took $y=y^{-1}=(1\ 2)$. 
Next I needed a permutation that doesn't commute with $(1\ 2)$, so I took $x=(1\ 3)$ and hoped for the best. [. . .]
That didn't pan out, so next I tried $x=(1\ 2\ 3)$. I found that 
$y^{-1}x=(1\ 2)(1\ 2\ 3)=(2\ 3)$ and $xy^{-1}=(1\ 2\ 3)(1\ 2)=(1\ 3)$. Now all I needed was a group $H$ containing $(2\ 3)$ but not $(1\ 3)$. For that I took the smallest group containing $(2\ 3)$, so $H=\langle(2\ 3)\rangle=\{(1),(2\ 3)\}$.
So now I had my counterexample: $H=\{(1),(2\ 3)\}$, $x=(1\ 2\ 3)$, $y=(1\ 2)$.
Finally, I double-checked:  $xH=(1\ 2\ 3)\{(1),(2\ 3)\}=\{(1\ 2\ 3),(1\ 2)\}$, and $yH=(1\ 2)\{(1),(2\ 3)\}=\{(1\ 2),(1\ 2\ 3)\}=\{(1\ 2\ 3)(1\ 2)\}=xH$, amd $xy^{-1}=(1\ 3)\notin H$, so everything is fine!
As for the second statement, I'm too lazy to start all over again, So I'll try using the same $x$ and $y$ as before, with a new $H$. I want $xy^{-1}\in H$ while $xH\ne yH$. Since $xy^{-1}=(1\ 3)$, I guess I'll take $H=\langle(1\ 3)\rangle=\{(1),(1\ 3)\}$. Is $xH=yH$? Well, $x\in xH$; is $x\in  yH$? Let's see, $yH=(1\ 2)\{(1),(1\ 3)\}=\{(1\ 2),(1\ 3\ 2)\}$, so $x\notin yH$ and $xH\notin yH$.
Well, that's how I did it. Doesn't seem all that hard. Where did you get stuck?
