# Thread: Cosets and Normal Subgroups

1. ## Cosets and Normal Subgroups

The group $D_8$ has two generators $x, y$ with the relations $x^4 = e, y^2 = e, xy = yx^3$. Let $H$ be the cyclic group generated by $x$. Write down the four elements in $H$. Write down the four elements in the left coset $yH$ and the four elements in the right coset $Hy$. Show that $yH = Hy$.

This is what I have so far.

$H = = \{e, x, x^2, x^3\}$
$yH = \{ye, yx, yx^2, yx^3\} = \{y, yx, yx^2, yx^3\}$
$Hy = \{ey, xy, x^2y, x^3y\} = \{y, x^2y, x^3y, yx^3\}$ (because $xy = yx^3$)

Now I am supposed to show that $yH = Hy$ and I cannot seem to manipulate the sets so they are equal. Thanks in advance for any help.

2. Ok, let's take the $Hy$ and see how we could get it to look like $yH.$ So the $y$'s are the same, no issue there. Try this for the $x^{2}y$ in $Hy:$

$x^{2}y=x(xy)=x(yx^{3})=(xy)x^{3}=yx^{3}x^{3}=yx^{6 }=yx^{4}x^{2}=yex^{2}=yx^{2},$ which is in $yH.$

I think you'll find that the other elements behave in a similar fashion. Can you finish from here?

3. Thank you so much for your help. I knew the answer was staring at me in the face. Thank you for pointing it out to me.

4. You're very welcome. Have a good one!