# Math Help - fun with dihedral groups

1. ## fun with dihedral groups

I have:

$D_n= \langle a,b | a^2 = b^2 =(ab)^2=1 \rangle$

A dihedral group can be generated by two distinct elements of order 2

(a) Show that this definition allows only one infinite group up to isomorphism and determine its centre.

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Is this right:

The only non-trivial rotation that commutes with all reflections is the rotation over $\pi$ . So $a$ is a rotation by $2 \pi/n$ and $b$ is a reflection.

Since $n \geq 3, \, a^{-1} \neq a$. Therefore,

$a^{r+1}b=a(a^rb)=(a^rb)a=a^{r-1}b$

So $a^2=1$ , which is a contradiction; thus, no element of the form $a^rb$ is in the center.

Similarly, if for $1 \leq s < n, a^sb=ba^s=a^{-s}b$ , then $a^{2s}=1$ , which is possible only if $2s=n$. Hence, $a^2$ commutes with $b$ iff $n=2s.$

So if $n=2s$ , the the center of $D_n$ is $\{1,a^s \}$; if $n$ is odd the center is $\{ 1 \}$.

2. Originally Posted by MichaelMath
I have:

$D_n= \langle a,b | a^2 = b^2 =(ab)^2=1 \rangle$

This is a presentation for the dihedral group $D_2$ , better known as the Klein group, the non cyclic

group of order 4

A dihedral group can be generated by two distinct elements of order 2

(a) Show that this definition allows only one infinite group up to isomorphism and determine its centre.

As already noted, the above is not an infinite group but a pretty finite one...

Tonio

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Is this right:

The only non-trivial rotation that commutes with all reflections is the rotation over $\pi$ . So $a$ is a rotation by $2 \pi/n$ and $b$ is a reflection.

Since $n \geq 3, \, a^{-1} \neq a$. Therefore,

$a^{r+1}b=a(a^rb)=(a^rb)a=a^{r-1}b$

So $a^2=1$ , which is a contradiction; thus, no element of the form $a^rb$ is in the center.

Similarly, if for $1 \leq s < n, a^sb=ba^s=a^{-s}b$ , then $a^{2s}=1$ , which is possible only if $2s=n$. Hence, $a^2$ commutes with $b$ iff $n=2s.$

So if $n=2s$ , the the center of $D_n$ is $\{1,a^s \}$; if $n$ is odd the center is $\{ 1 \}$.
.

3. Well then I dunno... I'll try and re-state it word for word:

Recall that the dihedral group $Dn$ is defined by $Dn=\langle a,b| a^2=b^2=(ab)^2=1 \rangle$ . More generally, a dihedral group is a group which can be generated by two distinct elements of order two.

Show that this definition allows for only one infinite group, up to isomorphism, and determine its centre. Find two reflection lines in the plane which generate an infinite dihedral group.

hmm...

4. Originally Posted by MichaelMath
Well then I dunno... I'll try and re-state it word for word:

Recall that the dihedral group $Dn$ is defined by $Dn=\langle a,b| a^2=b^2=(ab)^2=1 \rangle$ .

No. The above is NOT the general dihedral group $D_n$ but only the particular dihedral group $D_2=V=$the Klein group.
Please be very careful with: there is a little mistake you keep on doing...

Tonio

More generally, a dihedral group is a group which can be generated by two distinct elements of order two.

Show that this definition allows for only one infinite group, up to isomorphism, and determine its centre. Find two reflection lines in the plane which generate an infinite dihedral group.

hmm...
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5. Recall the dihedral group that is defined by $D_n= \langle a,b | a^2 = b^2 =(ab)^2=1 \rangle$. More generally, a dihedral group is a group which can be generated by two distinct elements of order two.

Show that this definition allows for only one infinite group, up to isomorphism, and determine its centre. Find two reflection lines in the plane which generate an infinite dihedral group.

Does this question make sense?

6. Oops. My bad. Now I know what I did wrong. I put $(ab)^2$ where there should have been $(ab)^n$ :

Recall that the dihedral group $D_n$ is defined by $D_n= \langle a,b \, |a^2=b^2=(ab)^n=1 \rangle$. More generally, a dihedral group is a group which can be generated by two distinct elements of order two.

Show that this definition allows for only one infinite group, up to isomorphism, and determine its centre. Find two reflection lines in the plane which generate an infinite dihedral group.

Any thoughts...

7. Originally Posted by MichaelMath
Oops. My bad. Now I know what I did wrong. I put $(ab)^2$ where there should have been $(ab)^n$ :

Recall that the dihedral group $D_n$ is defined by $D_n= \langle a,b \, |a^2=b^2=(ab)^n=1 \rangle$. More generally, a dihedral group is a group which can be generated by two distinct elements of order two.

Show that this definition allows for only one infinite group, up to isomorphism, and determine its centre. Find two reflection lines in the plane which generate an infinite dihedral group.

Any thoughts...
Basically, you should note that the group $D_n= \langle a,b \, |a^2=b^2=(ab)^n=1 \rangle$ is finite for all n, and so the infinite group you are looking for is the group $D_{\infty}= \langle a,b \, |a^2=b^2=1 \rangle$, which is the free product of $C_2$ with $C_2$, $C_2\ast C_2$. It looks like a free group, but generators have order 2.

Now, can you find an element which commutes with both of the generators?

8. Originally Posted by Swlabr
Basically, you should note that the group $D_n= \langle a,b \, |a^2=b^2=(ab)^n=1 \rangle$ is finite for all n, and so the infinite group you are looking for is the group $D_{\infty}= \langle a,b \, |a^2=b^2=1 \rangle$, which is the free product of $C_2$ with $C_2$, $C_2\ast C_2$. It looks like a free group, but generators have order 2.

Now, can you find an element which commutes with both of the generators?
I'm so tired... is it $e$?

9. Originally Posted by MichaelMath
I'm so tired... is it $e$?

Yes, but I really can't see an easy way to prove this without using the normal form for elements in a free product...

Tonio

10. I dunno... something about non-trivial elements can't commute so $e$.... ughh... I just found out what a triangle was last week.

11. Originally Posted by MichaelMath
I dunno... something about non-trivial elements can't commute so $e$.... ughh... I just found out what a triangle was last week.
As in...three points joined up with lines?...

What does every element of this group look like? Note that every generator, $a, b \in \langle a, b ; a^2=b^2=1\rangle$ is its own inverse so you never need to write $a^{-1}$ or $b^{-1}$.

Once you've found some form for your elements, take an arbitrary one, $w$. Can it be that $a^{-1}w^{-1}aw=aw^{-1}aw=1$? If so, what happens? Can it be that $au^{-1}awu=1$? If so, what happens?

Why then can't you get an element $w$ such that $aw^{-1}aw=1=bw^{-1}bw$?

12. Originally Posted by Swlabr
As in...three points joined up with lines?...

What does every element of this group look like? Note that every element is its own inverse so you never need to write things like $a^{-1}$.

I don't think this is true. For example, the inverse of $ab$ is $ba$ , not $ab$ again...

Tonio

Once you've found some form for your elements, take an arbitrary one, $w$. Can it be that $a^{-1}w^{-1}aw=aw^{-1}aw=1$? If so, what happens? Can it be that $au^{-1}awu=1$? If so, what happens?

Why then can't you get an element $w$ such that $aw^{-1}aw=1=bw^{-1}bw$?
.

13. Originally Posted by tonio
.
Sorry, that should read generator'. I'll edit it.

(Also, if every element was its own inverse the group would be abelian...)

14. Originally Posted by Swlabr
Sorry, that should read generator'. I'll edit it.

(Also, if every element was its own inverse the group would be abelian...)

Something that would be a true marvel taking into account that the infinite dihedral group is a free product of two non-trivial groups.

Anyway, is there any way to show that the infinite dihedral group is centerless without using it is a free product? I thought perhaps using

the fact that it is also the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}_2$ , with the obvious action of $\mathbb{Z}_2$ over $\mathbb{Z}$ , but

I don't think this would make things easier...

Tonio