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

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

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

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

gives a presentation for $\displaystyle D_{2n}$ 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 $\displaystyle {(ab)}^n$ = 1

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

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

Then $\displaystyle r^n$ = $\displaystyle {(s^{-1}b)}^n$ = $\displaystyle (a^{-1}b)^n$ = 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 $\displaystyle D_{2n}$

(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