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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.

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  • $\begingroup$ 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. $\endgroup$ May 17, 2011 at 15:33
  • $\begingroup$ 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. $\endgroup$ May 17, 2011 at 15:35
  • $\begingroup$ @Thomas, @Dane: Great, thanks guys! $\endgroup$
    – Altar Ego
    May 17, 2011 at 15:51
  • $\begingroup$ @Dane, that statement is exercise #3! $\endgroup$
    – Altar Ego
    May 17, 2011 at 19:19

1 Answer 1

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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

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