D4

• October 12th 2011, 03:56 AM
Bernhard
Orders of Elements of Dihedral Group D8 (Correction from earlier post)
Dummit and Foote (D&F) Section 1.2 Dihedral Groups, exercise 1(b) reads as follows:

Compute the order of each of the elements of $D_8$.

D&F give the following as one presentation of $D_{2n}$ :

$D_{2n}$ = < r,s | $r^n$ = $s^2$ = 1, rs = $sr^{-1}$>

Following this notation I proceeded to compute r, s and sr and (among other elements) was going to determine the order of sr. I proceeded as follows:

Given the set-up mentioned for the square - see attached pages from D&F on the Dihedral Groups - i defined r and s as follows:

r = $\left(\begin{array}{cccc} 1&2&3&4 \\ 2&3&4&1 \end{array}\right)$

s = $\left(\begin{array}{cccc} 1&2&3&4 \\ 1&4&3&2 \end{array}\right)$

Thus

sr = $\left(\begin{array}{cccc} 1&2&3&4 \\ 4&3&2&1 \end{array}\right)$

So from these permutations I calculated that

$(sr)^2$ = $\left(\begin{array}{cccc} 1&2&3&4 \\ 1&2&3&4 \end{array}\right)$ = 1

BUT!! from the relations given by D&F we have $s^2$ = 1 and so we can proceed thus to find $(sr)^2$

$(sr)^2$ = $s^2$ $r^2$ = 1. $r^2$ ????

Can anyone help?
• October 12th 2011, 06:36 AM
Deveno
Re: Orders of Elements of Dihedral Group D8 (Correction from earlier post)
in cycle notation, we have:

r = (1 2 3 4), s = (1 4)(2 3).

so sr = (1 3), a transposition, which is clearly of order 2.