# Thread: Cosets and Normal Subgroups

1. ## Cosets and Normal Subgroups

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

This is what I have so far.

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

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

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

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

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!