Proving Dn with Involutions: Group Representation Homework

JojoF
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Homework Statement


let n ≥ 2
Show that Dn = < a,b | a2, b2, (ab)n>

Homework Equations

The Attempt at a Solution


I see that a and b are involutions and therefore are two different reflections of Dn.

If we set set b = ar where r is a rotation of 2π/n

And Dn = <a,r | a2, rn, (ab)2 >

I am unsure how to write this as a clean proof... I would appreciate some ideas
 
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JojoF said:

Homework Statement


let n ≥ 2
Show that Dn = < a,b | a2, b2, (ab)n>

Homework Equations

The Attempt at a Solution


I see that a and b are involutions and therefore are two different reflections of Dn.

If we set set b = ar where r is a rotation of 2π/n

And Dn = <a,r | a2, rn, (ab)2 >

I am unsure how to write this as a clean proof... I would appreciate some ideas
What is ##D_n##? And no, do not answer "the dihedral group". The point is, that in order to show that ##A=B## where ##B## is explicitly given, you also have to say what ##A## is. Your description is only a letter.
 
fresh_42 said:
What is ##D_n##? And no, do not answer "the dihedral group". The point is, that in order to show that ##A=B## where ##B## is explicitly given, you also have to say what ##A## is. Your description is only a letter.
Of course you are right. I just copied an exercise verbatim. So let me restate the question.
Let n ≥ 2
Show that the dihedral group Dn is equal to <a,r | a2, rn, (ab)2> where the dihedral group is the group of symetries of a regular n-gon
 
Yes, but how are the symmetries given? The usual representation is by ##a## and ##r## and you have another representations by ##a## and ##b##. So one possibility is, to show that they are the same. If the symmetries are given by certain permutations of the vertices, then show that those permutations can be identified with ##a## and ##b## and obey the corresponding rules. If they are given by linear transformations, then proceed accordingly. So your definition is essential. If you only have "symmetries" then you have to first find a way to write those symmetries.
 
What if I write
Dn = {e,r,r2,..., rn-1,a,ar,ar2,..., arn-1} = <a,r | a2, rn, (ar)2 >

a set b = ra ⇒ r = ba-1 = ba

and we can therefore rewrite Dn = {e, ba, (ba)2, ... , (ba)n-1, a(ba), ..., a(ba)n-1 } = < a,b | a2, b2, (ab)n>
 
To show equality of sets, we need to show ##\subseteq## and ##\supseteq##, and for groups plus that the relations hold.
In case you already know ##D_n=\langle a,r\,|\,r^n=a^2=(ar)^2=1 \rangle## you are already almost done.

Now let ##D_n\,' =\langle a,b\,|\,a^2=b^2=(ab)^n=1 \rangle##.
##b=ra## yields ##b \in D_n##. Now we have to check the relations. We know ##b^2=1## so we have to show that ##(ra)^2=1## as well in ##D_n##. Now we do the same in the other direction: ##r=ba## since ##a^2=1##, so ##r \in D_n\,'##. We know ##r^n=1## and we have to show ##(ba)^n=1## in ##D_n\,'##. In the end we have shown, that the "new" elements ##b## and ##r## are within the other group and obey the relations there.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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