Presentation of $D_4$

Question

When one writes $D_4=\langle r,s\mid r^4=s^2=1,rsrs=1\rangle$ they are describing a quotient group. Let $S=\{s,r\}$ and $R=\{r^4,s^2,rsrs\}$. $$F_S=\langle r,s\rangle,\quad R^{F_S}=\{grg^{-1}\mid r\in R, g\in F_s\},\quad \langle R^{F_s}\rangle =N$$

and $D_4=F_S/N$. Is this correct?

Equivalently, they describe: $$F_1\overset{\partial_1}\to F_0\overset{\epsilon}\twoheadrightarrow D_4,$$ where $\partial_1:F_1\to \text{ker}(\epsilon)$ is a surjection.

Where then, I take it that $F_0=F_S$ and $F_1=N$?

Answer 1

If $F_S$ is the free group generated by $r$ and $s$, then yes, $D_4=F/N$. Just writing $F=\langle r,s\rangle$ doesn't make sense, because the notation $\langle r,s\rangle$ assumes that we are already working within some group.

Answer 2

You might prefer the short exact sequence $$1 \to N \overset{\partial_1} \hookrightarrow F_S\overset{\epsilon}\twoheadrightarrow D_4 \to 1$$ as this forces $\mathrm{im}\,\partial_1 = \ker \epsilon$, $\partial_1$ injective, and $\epsilon$ surjective.

The section on groups (same page) explains that this is equivalent to your first characterization; i.e., $D_4 \cong F_S/N$.