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As a follow-up to my previous question, the next two exercises state:

Use the given generators and relations to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then:

  • a) $rx = xr^{-1}$, and
  • b) $x$ has order 2, and hence $D_{2n}$ is generated by $s$ and $sr$, each with order 2.

Here, $D_{2n}$ has the "usual" presentation $$D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.$$

For part a), since $x$ is not a power of $r$, I have written $ x = sr^k$ for some $0 \leq k \leq n - 1$.

So then $rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}$, but this doesn't seem to be moving me in the right direction.

Should I be using induction on the power of $r$ in the element $x$ ?

Then for part b), I simply wrote $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1$, but I doubt that I have sufficiently justified my work.

From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble.

Thanks in advance for your help!

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  • $\begingroup$ You should use \langle instead of < and \rangle instead of > to denote "subgroup generated by". $\endgroup$ Commented May 17, 2011 at 19:50
  • $\begingroup$ @Arturo, thanks for the edits and info. I couldn't figure out that tex code! $\endgroup$
    – Altar Ego
    Commented May 17, 2011 at 19:58

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Why do you think you are not going in the right direction in part (a)? You've shown that $rx = sr^{k-1}$. Now, how much is $xr^{-1}$? Since $x=sr^k$, then $xr^{-1} = (sr^k)r^{-1} =\ldots$

Remember what you are trying to prove!

For part (b), you have to show that $$r^ks = sr^{-k}$$ which can be done by induction on $k$. Alternatively, why not use part (a)? You have $x=sr^k$, so then $$(sr^k)^2 = (sr^{k-1})r(sr^{k-1})r = sr^{k-1}sr^{k-1}r^{-1}r = (sr^{k-1})^2.$$ Repeating this until $k=0$ we get $x^2=1$.

Once you do that, you've correctly shown that $x$ has order dividing 2; now simply note that it cannot have order $1$ (it's not the identity, because the identity does lie in $\langle r\rangle$), so it has order exactly $2$.

But you aren't done, because you haven't addressed the second part. Can you show that $s$ and $sr$ generate $D_{2n}$? And that they each have order $2$? (The latter part is what you just finished; the former is easy: just show that you can obtain all elements from a generating set by using $s$ and $sr$).

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  • $\begingroup$ Oh man, (a) was a lot easier than I realized! I will work on the end of (b). In the meantime, should I somehow edit my question to include this progress? $\endgroup$
    – Altar Ego
    Commented May 17, 2011 at 20:17
  • $\begingroup$ @Altar: You can; if you do, I would suggest adding it (say, below a horizontal line added with <hr/>). That way it won't seem like the answers are going over stuff you already knew. $\endgroup$ Commented May 17, 2011 at 20:20
  • $\begingroup$ @Altar: Now that you have the reputation for it, don't forget to start voting up answers that you find helpful/interesting (and questions others post that you find interesting); also, eventually you'll want to accept answers to your questions, if you are satisfied. Check out the FAQ for more info on all of that. $\endgroup$ Commented May 17, 2011 at 20:32
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For (a), I reckon induction is the way to go. You have the case $k=0$ in the group presentation.

For (b), your argument looks sound. The only part you might want to think about is justifying $r^ks=sr^{-k}$, but that's an almost immediate consequence of (a).

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