1. ## Groups - presentations

Dummit and Foote exercise 17 Section 1.2 Dihedral Groups reads as follows:

Let $\displaystyle X_{2n}$ be the group whose presentation is

$\displaystyle X_{2n}$ = <x,y | $\displaystyle x^n$ = $\displaystyle y^2$ = 1, xy = y$\displaystyle x^2$>

Show that if n = 3k, then $\displaystyle X_{2n}$ has order 6, and it has the same generators and relations as $\displaystyle D_6$ when x is replaced by r and y is replaced by s.

Can anyone help with this problem?

Can anyone with some insight into this problem guess what D&F meant by n = 3k - would it mean n is a multiple of 3 - that is k is a positive integer? I presume they mean this.

2. ## Re: Groups - presentations

yes, that is what is meant. the trick is to prove that if n = 3k, then k = 1.

note that $\displaystyle x = xy^2 = (xy)y = yx^2y$ . so

$\displaystyle yx = y(yx^2y) = x^2y$ ,

and $\displaystyle yx^2 = (yx)x = (x^2y)x = x^2(yx) = (x^2)(x^2y) = x^4y$.

but $\displaystyle yx^2 = xy$, so $\displaystyle x^4y = xy \implies x^3 = 1$.

it should be downhill from here.