# Groups - presentations

• Oct 15th 2011, 03:24 AM
Bernhard
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.
• Oct 15th 2011, 03:38 PM
Deveno
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.