# Math Help - Presentations of D2n

1. ## Presentations of D2n

Dummit and Foote in Section 1.2 Dihedral Groups give the following presentation of $D_{2n}$

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

Exercise 7 at the end of D&F Section 1.2 is as follows:

Show that <a,b | $a^2$ = $b^2$ = ${(ab)}^n$ = 1>

gives a presentation for $D_{2n}$ in terms of the two generators a = s and b = sr. [Show that the relations for r and s follow from the relations for a and b and conversely that the relations for a and b follow from those for r and s]

I was able (to my satisfaction anyway) to show all relations followed except for assuming the relations for r and s and trying to show that ${(ab)}^n$ = 1

In this case I, of course, started with $r^n$ = $s^2$ = 1, rs = s $r^{-1}$

and then put s = a and r = $s^{-1}$b = $a^{-1}$b

Then $r^n$ = ${(s^{-1}b)}^n$ = $(a^{-1}b)^n$ = 1

But I can get no further ... can anyone help?

A further question is this if

(1) we assume the presentation in terms of r and s does as D&F assert generates the whole group $D_{2n}$

(2) another presentation such as the one above based on a and b is such that the relations involved can be derived from the first

- then we can conclude that the new presentation can be used to derive the whole group? Is that right?

Peter

2. ## Re: Presentations of D2n

given: r^n = s^2 = 1, rs = sr^-1, and a = s, b = sr

prove: (ab)^n = 1.

so we must show that (srs)^n = 1. note that since s^2 = 1, s = s^-1. so we are showing that (srs^-1)^n = 1.

but (srs^-1)^n = sr^ns^-1 = ss^-1 (since r^n = 1)

= 1.

(1 ) you can prove the presentation does generate D2n (how many distinct elements can we have?. which elements are the rotations, and which are the reflections?)

(2) this is what you presumably just showed. since each presentation is derivable from the other, the two groups presented are isomorphic (showing that two isomorphic groups need not have the same presentation, which is part of a much larger problem within group theory).

Thanks