Proving that D_{2n} has generators s, rs

  • Thread starter Mr Davis 97
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In summary: But ##r\cdot s\in\langle s,rs\rangle = G##, so by closure, ##r \in G##. The other direction is similar.
  • #1
Mr Davis 97
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Homework Statement


Show that the subgroup of ##D_{2n}## generated by the set ##\{s, rs \}## is ##D_{2n}## itself. (i.e. show that ##\{s, rs\}## is another set of generators different than ##\{r,s\}##).

Homework Equations

The Attempt at a Solution


It's not clear to me what exactly I need to do. Just as some background, I have it defined that if ##G## is a group and ##X \subseteq G## then ##\langle X \rangle = \{x^{j_1}_1x^{j_2}_2 \dots x^{j_m}_m ~|~ m, j_1, \dots, j_m \in \mathbb{Z}, x_1 \dots, x_m \in X\}##. Will this definition be useful to me? Using the definition do I need to show that ##\langle s, rs \rangle = \langle r, s \rangle##?
 
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  • #2
Mr Davis 97 said:

Homework Statement


Show that the subgroup of ##D_{2n}## generated by the set ##\{s, rs \}## is ##D_{2n}## itself. (i.e. show that ##\{s, rs\}## is another set of generators different than ##\{r,s\}##).

NOTE: the title is wrong. It's rs, not sr.
Corrected.

Homework Equations

The Attempt at a Solution


It's not clear to me what exactly I need to do. Just as some background, I have it defined that if ##G## is a group and ##X \subseteq G## then ##\langle X \rangle = \{x^{j_1}_1x^{j_2}_2 \dots x^{j_m}_m ~|~ m, j_1, \dots, j_m \in \mathbb{Z}, x_1 \dots, x_m \in X\}##. Will this definition be useful to me? Using the definition do I need to show that ##\langle s, rs \rangle = \langle r, s \rangle##?
Yes, but isn't it pretty obvious? ##D_{2n}=\langle r,s\,|\,r^n=s^2=(rs)^2\rangle## now can you express ##r=w(s,rs)## as a word with letters ##s## and ##rs\,?##
 
  • #3
fresh_42 said:
Yes, but isn't it pretty obvious? ##D_{2n}=\langle r,s\,|\,r^n=s^2=(rs)^2\rangle## now can you express ##r=w(s,rs)## as a word with letters ##s## and ##rs\,?##
Well ##(rs)s = rs^2 = r##. Is the fact that ##(rs)^0s = s## also important? Am I showing that since every element ##x## in ##D_{2n}## can be written as ##x = s^ir^j##, and we have that ##s^ir^j = s^i[(rs)s]^j##, then ##\langle s, rs \rangle = D_{2n}##?
 
  • #4
Mr Davis 97 said:
Well ##(rs)s = rs^2 = r##. Is the fact that ##(rs)^0s = s## also important? Am I showing that since every element ##x## in ##D_{2n}## can be written as ##x = s^ir^j##, and we have that ##s^ir^j = s^i[(rs)s]^j##, then ##\langle s, rs \rangle = D_{2n}##?
No, ##(rs)^0s=s## is nonsense, as you could add as many ##e## as you want. Where would you stop? We usually use shortened presentations of words, at least as far as possible and if nothing else is intended, e.g. with constructions where ##g\cdot g^{-1}## must be inserted to demonstrate something. Here we have ##\langle r,s \rangle \supseteq \langle s,rs \rangle## which is obvious, and for the other direction, we need to make sure that ##r,s \in \langle s,rs \rangle##. But ##s## is per definition part of it and ##r=(rs)s## by associativity and identity element.
 
  • #5
fresh_42 said:
No, ##(rs)^0s=s## is nonsense, as you could add as many ##e## as you want. Where would you stop? We usually use shortened presentations of words, at least as far as possible and if nothing else is intended, e.g. with constructions where ##g\cdot g^{-1}## must be inserted to demonstrate something. Here we have ##\langle r,s \rangle \supseteq \langle s,rs \rangle## which is obvious, and for the other direction, we need to make sure that ##r,s \in \langle s,rs \rangle##. But ##s## is per definition part of it and ##r=(rs)s## by associativity and identity element.
So would the proof go something like this?:

##\langle s, rs \rangle = D_{2n}## if and only if ##\langle r, s \rangle = \langle s, rs \rangle##, so it is sufficient to show that the latter equality holds. The containment ##\langle r, s \rangle \subseteq \langle s, rs \rangle## holds: since ##r, s \in \langle s, rs \rangle## (cleary this is true for ##s##, and also ##r=(rs)s##), every element in ##\langle r, s \rangle## is also an element of ##\langle s, rs \rangle##. The containment ##\langle s, rs \rangle \subseteq \langle r, s \rangle## is clear, because since ##s,rs \in \langle r, s \rangle## every element in ##\langle s, rs \rangle## is also an element in ##\langle r, s \rangle##. Therefore ##\langle r, s \rangle = \langle s, rs \rangle = D_{2n}##.
 
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  • #6
Mr Davis 97 said:
So would the proof go something like this?:

##\langle s, rs \rangle = G## if and only if ##\langle r, s \rangle = \langle s, rs \rangle##, so it is sufficient to show that the latter equality holds. The containment ##\langle r, s \rangle \subseteq \langle s, rs \rangle## holds: since ##r, s \in \langle s, rs \rangle## (cleary this is true for ##s##, and also ##r=(rs)s##), every element in ##\langle r, s \rangle## is also an element of ##\langle s, rs \rangle##. The containment ##\langle s, rs \rangle \subseteq \langle r, s \rangle## is clear, because since ##s,rs \in \langle r, s \rangle## every element in ##\langle s, rs \rangle## is also an element in ##\langle r, s \rangle##. Therefore ##\langle r, s \rangle = \langle s rs \rangle = D_{2n}##.
I would have written it differently, if written as a correct proof, but the arguments are correct. E.g. I would written the groups with relations, because otherwise they are different groups, and instead of so many words, I'd use a bit more formula instead, so my proof would look like:

Show: ##G:=\langle s,rs\,|\,s^2=(rs)^2=r^n=1\rangle = \langle r,s\,|\,r^n=s^2=(rs)^2=1\rangle =:D_{2n}\,.##
Proof: ##G \subseteq D_{2n}## by definition, as well as ##s \in G##. So it remains to show ##G \ni r=r\cdot 1 =r\cdot s^2=(r\cdot s)\cdot s\,.##
 

1. How do you prove that D2n has generators s and rs?

The proof for this involves showing that s and rs can generate all the elements of D2n. This can be done by demonstrating that any element of D2n can be written as a product of s and rs, and that the generators satisfy the defining relations of D2n.

2. Why are s and rs considered generators of D2n?

In D2n, s represents a reflection and rs represents a rotation. Together, they can generate all the symmetries of a regular n-gon by reflecting and rotating it. This is why they are considered generators of D2n.

3. Can you provide an example of how s and rs generate elements of D2n?

For a regular hexagon, s can be a reflection across the horizontal axis and rs can be a rotation of 60 degrees counterclockwise. By combining these two actions, we can generate all the symmetries of the hexagon, including rotations of 120, 180, 240, 300 degrees and reflections across the other axes.

4. Is it possible for D2n to have more than two generators?

Yes, it is possible for D2n to have more than two generators. In general, D2n can have up to n generators. However, it has been proven that s and rs are the minimum number of generators needed for D2n.

5. How does the proof for D2n having generators s and rs differ from other groups?

The proof for D2n having generators s and rs is specific to this group and cannot be applied to other groups. Each group has its own set of defining relations and generators, so the proof will vary depending on the group being studied.

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