Groups - presentations

**Bernhard** · October 15th 2011, 03:24 AM

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

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

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

Show that if n = 3k, then $X_{2n}$ has order 6, and it has the same generators and relations as $D_6$ 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.

**Deveno** · October 15th 2011, 03:38 PM

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

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

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

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

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

it should be downhill from here.

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