Isomorphism question

• May 22nd 2010, 12:58 PM
aabsdr
Isomorphism question
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.
• May 22nd 2010, 08:20 PM
TheArtofSymmetry
Quote:

Originally Posted by aabsdr
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.

$\displaystyle D_{2n}$ of order 2n have presentations

1. $\displaystyle <r, s |r^n=1, s^2=1, sr=r^{-1}s>$,
2. $\displaystyle <x, y | x^2 = y^2 = {(xy)}^n=1>$.

2 is denoted as $\displaystyle I_2(n)$, implying that $\displaystyle D_{2n}$ 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 $\displaystyle <a, b | a^2 = b^2 = {(ab)}^2=1>$, which is $\displaystyle D_4$.

I'll leave it to you to covert the presentation 1 into 2 and verify that they are indeed equivalent.
• May 23rd 2010, 01:23 PM
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.
• May 23rd 2010, 08:53 PM
aabsdr
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>?
• May 24th 2010, 12:34 AM
Swlabr
Quote:

Originally Posted by aabsdr
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 $\displaystyle x^iy^j$ for $\displaystyle i, j \in \{0, 1, 2, 3\}$. What you need to do is find the values of $\displaystyle i,j$ such that $\displaystyle x^iy^jx^ay^b = x^ay^ba^iy^j$ for all $\displaystyle a, b$. So just plug it in...

Alternatively, noting that you are in $\displaystyle D_{2n}$ you can do the same trick. And yes, your group will have a center of order 2 in this case as $\displaystyle D_{2n}$ has a center of order 2 if and only if $\displaystyle n$ is even. Otherwise, it has trivial center.
• May 24th 2010, 02:10 AM
TheArtofSymmetry
Quote:

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.

In $\displaystyle G/<y^2>$, simply plug $\displaystyle y^2=1$ into your presentation of G.

Then, the original presentation reduces to $\displaystyle <x, y|x^2 = y^2 = xyxy^{-1}=1>=<x, y|x^2 = y^2 = xyxy=1>$, where $\displaystyle y=y^{-1}$ since $\displaystyle y^2=1$. 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.
• May 24th 2010, 02:40 AM
Swlabr
Quote:

Originally Posted by TheArtofSymmetry
In $\displaystyle G/<y^2>$, simply plug $\displaystyle y^2=1$ into your presentation of G.

(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, $\displaystyle \langle X ; R=1 \rangle :\cong F(X)/<<R>>$ where $\displaystyle F(X)$ is the free group over the alphabet $\displaystyle X$ and $\displaystyle <<R>>$ is the normal subgroup generated by the elements of $\displaystyle R$.)