Originally Posted by

**aabsdr** Here is what I have:

since xyxy^-1=e, we know that yxy^-1=x^-1=x^3, so we know that <x) is normal. So G= <x> union y<x> union y^2<x> union y^3<x> and |G|<= 16.

Lets assume that |G|=16.

So |G|/|<y^2>|=8, thus |G|/|<y^2>|>= |D4|.

D4=<a,b|a^4=b^2=(ab)^2=e, or ab=ba^3>.

Let w=x<y^2> , z=y<y^2> and q=e<y^2> (where q is are identity element in G/<y^2>.

Need to show that G/<y^2>= <w,z|w^4=z^2=(wz)^2=e>.

1. Since x and y are generates in G, then w and z are generates in G/<y^2>. I am not sure if this is right, could someone explain this to me? (I need to show that w and z generate G/<y^2>).

2. w^4=e since (x<y^2>)^4=x^4<y^2>=e<y^2>=q.

3. z^2=e since (y<y^2>)^2=y^2<y^2>=e<y^2>=q.

4. (wz)^2=(wz)(wz)=e. Since wz=(z^-1)(w^-1), then wz=xy<y^2> and z^-1w^-1=y^3x^3<y^2>=yx^3<y^2>. Thus xy<y^2>=yx^3<y^2>.

Therefore, G/<y^2> is isomorphic to D4.