# Thread: Generating a subgroup of D(2n)

1. ## Generating a subgroup of D(2n)

First, just to make sure we are all working with the same notation: D(2n) is the Dihedral group order 2n, where (in a physical representation) n is the number of points of a regular plane figure, r is the usual rotation of the plane element rotating the figure by one point in the clockwise direction, and s is the inversion element. The presentation I'm using is <r, s| r^n = s^2 = 1, rs = s r^{-1} >.

Okay. Here's my problem. I am constructing a set X(2n) from D(2n) by selecting all the x elements from D(2n) that are not powers of r and have the property rx = xr^{-1}. The problem is that my text claims that X(2n) is a group. But the identity does not belong to X(2n): $re = er \neq er^{-1}$ unless |r| = 2, which is not general. Not only that, the multiplication in X(2n) is not closed as it contains elements of the sort s(sr) = (ss)r = er = r, which is not in X(2n) by definition.

I could take this all as a typo, but the very next problem in the set does the same thing with another kind of sub"group" of D(2n).

Am I going bonkers?

-Dan

2. ## Re: Generating a subgroup of D(2n)

You are not going bonkers. The set $X(2n)$ you are describing is not a group. The smallest subgroup of $D(2n)$ containing $X(2n)$ is $D(2n)$. But, $X(2n) = \{sr^k \mid k \in \mathbb{Z}, 0\le k < n\}$. So, it is just the group of rotations multiplied by $s$. So, the set $sX(2n)$ is a group.