Trying to prove any relation between elements of a group can be derived from the relations in the presentation of the group

Question

In Abstract Algebra by Dummit and Foote on pg 26, it says:

In $D_{2n}$ we have relations: $r^{n}=1, s^{2}=1,$ and $rs=sr^{-1}$. Moreover, in $D_{2n}$ these three relations have the additional property that any other relation between elements of the group may be derived from these three (this is not immediately obvious; it follows from the fact that we can determine exactly when two group elements are equal by using only these three relations.)

I'm still a bit confused on how to prove any relation between elements of the group may still be derived from these 3 basic equations. Note: I've just started learning abstract algebra so the simpler the proof the more I would appreciate it. Thanks!

Note: similar question was asked here: Relations in Group Presentation but I still don't see how to approach the proof I want.

Answer 1

There's some subtlety here, and it's not at all an obvious fact. I'll walk you through the ideas involved, but I'll leave it up to you to verify some details.

Remember what a presentation means. When I write

$$D_8 = \langle r,s ~|~ r^4 = 1, s^2 = 1, srs = r^{-1} \rangle$$

what I mean is that $D_8 \cong F_2 / N$ where $F_2$ is the free group with generators $r$ and $s$, and $N$ is the "normal closure" of $\{r^4, s^2, srsr\}$. That is, $N$ is the smallest normal subgroup containing those elements.

When we quotient, we are imposing relations on the free group. By this I mean we are forcing certain products to be $1$. Why should all of these relations be written as a product of the relations $r^4 = 1$, $s^2 = 1$, and $srsr = 1$? Well for that we need to understand what the normal closure of a set is.

Definition:

If $S$ is a subset of a group $G$, the normal closure $\langle S^G \rangle$ is the smallest normal subgroup containing $S$. That is, $$\langle S^G \rangle = \bigcap \{ H \trianglelefteq G ~|~ S \subseteq H \}.$$

Theorem (exercise):

$g \in \langle S^G \rangle$ if and only if $g$ is a product of elements of the form $xsx^{-1}$ for $x \in G$ and $s \in S$. That is, if and only if $g$ is a product of conjugates of elements of $S$.

Now we see why the claim is true. Since there are no relations in $F_2$, all relations in $D_8$ come from $N$. But all the relations in $N$ are products of conjugates of the three listed relations. This is the sense in which those relations generate all of them. Of course, there is nothing special about $D_8$. This idea works for any presentation of a group.

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I hope this helps ^_^