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

  • #1
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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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Answers and Replies

  • #2
12,652
9,172

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
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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
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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
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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
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9,172
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\,.##
 

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