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.
of order 2n have presentations
1. ,
2. .
2 is denoted as , implying that can also be generated by two reflections instead of one rotation and one reflection. You can find this idea in the context of Coxeter groups.
If you just play a little bit of your question, you will reduce your presentation to , which is .
I'll leave it to you to covert the presentation 1 into 2 and verify that they are indeed equivalent.
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.
how would I find the center for this group if the groups order was 16? I know that since xyxy^-1=e that xy=yx^3, so x=yx^3y^-1, so than x^2y=xyx^3=yx^6=yx^2. So x^2 and e is in the center. How would I show that nothing else is in the center besides <x^2>?
Looking at the presentation you give, the third relation gives you something which looks like commutativity. Essentially, it allows you to write every element in the form for . What you need to do is find the values of such that for all . So just plug it in...
Alternatively, noting that you are in you can do the same trick. And yes, your group will have a center of order 2 in this case as has a center of order 2 if and only if is even. Otherwise, it has trivial center.
In , simply plug into your presentation of G.
Then, the original presentation reduces to , where since . It is exactly the presentation of D_4 of order 4 (See my previous reply).
To convert the presentation 1 into presentation 2 in my previous reply, substitue s for x and sr for y. Then you will see that both are equivalent presentations.
(It should perhaps be pointed out that relations in a presentation, homomorphic images and quotients are all equivalent. Thus, putting a relation into your presentation is the same as taking a quotient.
This is because, formally, where is the free group over the alphabet and is the normal subgroup generated by the elements of .)