Isomorphism question with D4

[tyrannosaurus]

Homework Statement

Let G=<x, y|x^4=y^4=e, xyxy^–1=e>. Show that |G|≤16. Assuming |G|=16, show G/<y^2> is isomorphic to D4.

Homework Equations

The Attempt at a Solution

Here is what I have:
since xyxy^-1=e, we know that yxy^-1=x^-1=x^3, so we know that x is a conjugate and partions G. So G= <x> union y<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.

 

[mighty2000]

Your solution looks good overall! Here are a few clarifications and suggestions:

1. You are correct that w=x<y^2> and z=y<y^2> are generators for G/<y^2>. This is because any element in G/<y^2> can be written as (x^i)(y^j)<y^2> for some integers i and j. Since x and y are generators for G, this means that w and z can generate any element in G/<y^2>.

2. You can also show that w^4=q and z^2=q using the same reasoning as in your attempt. Since x^4=y^4=e, this means that (x<y^2>)^4=(y<y^2>)^4=q.

3. In step 4, you are correct that wz=xy<y^2>=yx^3<y^2>. However, to show that (wz)^2=e, you need to show that (wz)(wz)=e. This follows from the fact that wz=xy<y^2> and z^-1w^-1=yx^3<y^2>, as you mentioned. So (wz)(wz)=xy<y^2>yx^3<y^2>=e.

Overall, your solution is clear and correct. Great job!