Thread: D4

**Bernhard** · October 12th 2011, 02:56 AM

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?

**Deveno** · October 12th 2011, 05:36 AM

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.

your mistake is in your calculation of (sr)^2 algebraically.

it is NOT the case that (sr)^2 = s^2r^2. typically, that is only true in abelian groups.

what you should have is this:

(sr)^2 = (sr)(sr) = s(rs)r (by associativity)

= s(sr^-1)r (from rs = sr^-1)

= (ss)(r^-1r) = (1)(1) = 1.

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Orders of Elements of Dihedral Group D8 (Correction from earlier post)

Re: Orders of Elements of Dihedral Group D8 (Correction from earlier post)