The presentation of the dihedral group $D_{2n}$

Question

We know that $D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$ From Dummit & Foote, any other relations of $r,s$ can be deduced from the relations given in the presentation. The book claims that this is true because we can tell whether two elements are equal using the given relations. Why being able to tell whether two elements are equal using the given relations implies that any other relations of $r,s$ can be deduced from the relations given in the presentation.

Answer 1

One way to think of this is to realise that in $D_{2n}$, you have two kinds of objects that interact - the rotations $r$ and the reflections $s$.

The defining property of the rotation is that $r^n = 1$, that is, after $n$ rotations, you're back to identity.

Similarly, the defining properly of a reflection is $s^2 = 1$, which tells you that two reflections are the same as no reflection.

The next component you need is to relate rotations and reflections:

$$ rs = sr^{-1} $$

that is, a rotation $r$ followed by a reflection $s$ is the same as the reflection $s$ followed by ($360^\circ$ - $r$).

Now, let us assume that we are given some element of the dihedral group

$$ d = r_0^{i_0}s_0^{j_0} \cdot r_1^{i_1}s_1^{j_1} \ldots $$

we can issue the requirement that $i_k < n$, since all $i_k$ are the coefficients of the rotations $r_k$, and we can collapse large rotation coefficients to become less than $n$ by using $r^n = 1$

Similarly, we can require that $j_k = 0, 1$ since $s^2 = 1$

Now, to reduce $d$, we can use the fact that $rs = sr^{-1}$ like so:

$$ d = r_0^{i_0}s_0^{j_0} \cdot r_1^{i_1}s_1^{j_1} \ldots \\ \text {swizzling $r_0$, $s_0$} \\ \\ d = s_0^{j_0}r_0^{-i_0} \cdot r_1^{i_1}s_1^{j_1} \ldots \\ d = s_0^{j_0}r^{i_1 - i_0}s_1^{j_1} \ldots \\ \text{swizzling $r$, $s_0$ } \\ d = r^{-(i_1 - i_0)}s_0^{j_0}s_1^{j_1} \ldots \\ d = r^{i_0 - i_1}s^{j_0 + j_1} $$

If that's not clear, we first took $r_0^{i_0} s_0^{j_0} \to s_0^{j_0} r_0^{-i_0}$.

We then composed $r_0^{-i_0} r_1^{i_1} \to r^{i_1 - i_0}$, which gave us one rotation.

Next, we chose to swizzle around $s_0^{j_0}r^{i_1 - i_0} \to r^{-(i_1 - i_0)}s_0^{j_0}$, so we could compose the $s_0$ and $s_1$ together in the next step.

Finally, we compose $s_0^{j_0} s_1^{j_1} \to s^{j_0 + j_1}$.

Clearly, we have reduced a 2-term $r_0^{i_0}s_0^{j_0} r_1^{i_1} s_1 ^{j_1} \to r^{i_0 - i_1} s^{j_0 + j_1}$

This process can be extended to $n$ terms, allowing us to collapse a large presentation of a single element into one $d = r^x s^y$.