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

## 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 $\{r,s\}$).

## 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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fresh_42
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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 $\{r,s\}$).

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

## 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 $rs\,?$

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 $\langle s, rs \rangle = D_{2n}$?

fresh_42
Mentor
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.

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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fresh_42
Mentor
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 $G \ni r=r\cdot 1 =r\cdot s^2=(r\cdot s)\cdot s\,.$