Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


I am working through "Abstract Algebra" by Dummit & Foote. Exercise 1.2.1 states:

Compute the order of each of the elements in ... $D_6$, $D_8$, and $D_{12}$.

I have found that: $$\begin{align*} D_6 &= \{ 1, r, r^2, s, sr, sr^2 \},\\ D_8 &= \{ 1, r, r^2, r^3, s, sr, sr^2, sr^3 \}, \qquad\text{and}\\ D_{12} &= \{ 1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5 \}. \end{align*}$$

In these descriptions, $r$ is rotation of $\frac{2\pi}{n}$ radians, and $s$ is reflection through vertex "1" and the origin.

Also, this book uses $D_{2n}$ for the dihedral group of rigid motions of the regular $n$-gon (that is, the index is the order of the group).

Determining the orders of powers of $r$ is easy (and $|s| = 2$), but I get confused when trying to compute, say, $|sr^5|$.

I think the key is the statement: $r^is = sr^{-i}$ for $0 \leq i \leq n$.

Then we would have $$ (sr^5)(sr^5) = s(r^5s)r^5 = s(sr^{-5})r^5 = (ss)(1) = s^2 = 1 $$

So $|sr^5| = 2$ ?

Am I doing this correctly? Is there a better way?
Thanks.

share|improve this question
    
Hint: you can use the fact that a dihedral group is a group generated by two involutions. For instance, the group $D_{2n}$ has presentation $\langle s,t \mid s^2=t^2=(st)^n = 1 \rangle$. This means that $s$ and $t$ are both reflections through lines whose angle is $\pi/n$. Now any element of $D_{2n}$ is of the form $ststst\ldots st$ or so. Now it should be much easier to compute products. –  Thomas Connor May 17 '11 at 15:33
    
Yes, that's exactly right. By the same argument, all the elements outside of the cyclic subgroup generated by $r$ (elements of the form $sr^i$) will have order 2. –  Dane May 17 '11 at 15:35
    
@Thomas, @Dane: Great, thanks guys! –  Altar Ego May 17 '11 at 15:51
    
@Dane, that statement is exercise #3! –  Altar Ego May 17 '11 at 19:19
add comment

1 Answer

up vote 5 down vote accepted

Yes, you are perfectly right. If you think at the problem geometrically, probably you will get a sharper picture of the situation.

Every element of the form $sr^k$ is in fact a reflection around some axes, thus it's very natural to expect it has order 2. Your proof is correct, indeed; just replace the $5$ with a $k$ and you've done!

bye

share|improve this answer
    
Thank you for your reply. –  Altar Ego May 17 '11 at 19:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.