# Commutator subgroup of a dihedral group.

I have a few questions concerning an example of the commutator subgroups in the dihedral group. This example is found on pg.171 of Abstract Algebra by Dummit and Foote.

Let $D_{2n}=\langle r,s |r^n=s^2=1, s^{-1}rs=r^{-1}\rangle$. Since $[r,s]=r^{-2}$ we have that $\langle r^{-2} \rangle = \langle r^2 \rangle \le D'_{2n}$. Furthermore, $\langle r^2\rangle \trianglelefteq D_{2n}$ and the images of $r$ and $s$ in $D_{2n} / \langle r^2 \rangle$ generate this quotient.

What exactly is meant by the image of $r$ and $s$?

They (is this referring to $r$ and $s$?) are commuting elements of order $\le 2$ (I know $s$ is of order $2$ but $r$ is of order $n$??) so the quotient is abelian and $D'_{2n} \le \langle r^2 \rangle$. (I thought the quotient was abelian due to the already established properties of $\langle r^2 \rangle$ i.e it is normal and a subgroup of the commutator subgroup.)

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I think you have confused the commutator subgroup with the quotient. The quotient is (isomorphic to the group) formed by adding the relation $g=1$ for every generator of the subgroup. In this case, the only generator is $r^2$, so we get:

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

$$=<r, s | s^2=r^{n \text{mod} 2}=rsrs=1>$$

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I am not sure I understand. Isn't the quotient formed by the commutator $<r^2>$? I understand that it is also a generator but why is this of any significance. –  fafddf Dec 5 '13 at 3:29
$<r^2>$ is not a commutator --- note the angle brackets. It is a subgroup. It is generated by the (only interesting) commutator $[r, s] = r^2$. We are forming the quotient by that subgroup. –  apt1002 Dec 5 '13 at 4:16
The definition of the quotient is that it is a group whose elements are cosets of $<r^2>$. In my answer, I represented each coset by one of its elements. To make the trick work I added the relation $r^2=1$. Am I helping? –  apt1002 Dec 5 '13 at 4:21