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I’m reading up on presentations of a group. In here, the example is the dihedral group $D_8$, with generators rotation $r$ of order $8$, and flip $f$ of order $2$. In the construction of the presentation of $D_8$ from the free group $F$ generated by those two letters $r$ and $f$, it’s fairly obvious why $r^8 = 1$ and $f^2 = 1$ are the necessary relations. But what about $rfrf = 1$? Where does it come from? What happens if we drop it from the presentation, i.e. if we just have $\langle r,f \mid r^8 = 1, f^2 = 1 \rangle$?

Also, just to be sure, when the article says that $D_8 \cong F/N$, does the argument implicitly involves a homomorphism $\phi: F \to D_8$, with $N$ as the kernel of $\phi$?

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    $\begingroup$ $rfrf=1$ comes from the geometric fact that if you apply $rfr$ to an octagon, the result is the same as $f$, so that $rfr=f$, or $rfrf=1$. $\endgroup$ – rogerl Mar 3 at 14:46
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The group $\langle r,f \mid r^8 = 1, f^2 = 1 \rangle$ is infinite, and the dihedral group $D_8 = \langle r,f \mid r^8 = 1, f^2 = 1, (rf)^2 = 1 \rangle$ is finite. And the comment of @rogerl explains the geometric reason for that relation $(rf)^2=1$.

To get to the heart of the matter, here's how to prove that for any word in the free group $F = \langle r,f \rangle$, it's image in $D_8$ is equal to the image of an element of the form $r^a f^b$ where $a \in \{0,1,2,3,4,5,6,7\}$ and $b \in \{0,1\}$. Start with any word in $r$, $r^{-1}$, $f$, $f^{-1}$. By combining consecutive $f$'s into one power of $f$, and similarly for $r$, one obtains a word that alternates between powers of $f$ and powers of $r$. Using $f^2=1$, one can replace each power of $f$ with either the empty word or $f$ itself. Using $frf=r^{-1}$, which together with $f=f^{-1}$ implies that $fr^af=r^{-a}$ one can then reduce the number of $f$'s by $2$, eventually reaching a word of one of the forms $r^a f r^b$ or just $r^a$. To simplify the first form further, using $fr=r^{-1}f$, one can move the $f$ to the end, and one gets $r^a f$ or just $r^a$. Using $r^8=1$, one can reduce $a$ modulo $8$. Thus, group element is equal to one of the form $r^a f^b$ where $a \in \{0,1,2,3,4,5,6,7\}$ and $b \in \{0,1\}$.

Regarding your last paragraph, what you write is essentially correct, and moreover it holds in great generality. To say that a group $G$ has a presentation $G \cong \langle s_i \mid r_j\rangle$ means that there is a surjective homomorphism $\phi : F \langle s_i \rangle \to G$ defined on the free group $F = F\langle s_i \rangle$ with free basis $\{s_i\}$, such that the kernel $N$ of $\phi$ is normally generated by the subset $\{r_j\} \subset F$, and therefore $G \cong F / N$.

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  • $\begingroup$ Thank you for the answer. What’s still not clear to me is why that particular relation $rfrf = 1$ (together with $r^8 = 1$ and $f^2 = 1$) uniquely determines $D_8$. From what I understand, $F$ is a collection of reduced words built from $r$ and $f$, with “reduced” defined via the two relations $r^8 = 1$ and $f^2 = 1$. It’s also easy to check that $rfrf = 1$ holds for $D_8$, so any subset of $F$ isomorphic to $D_8$ must also satisfies this added relation. But how do we know that there’s no other relations that need to be included? $\endgroup$ – ensbana Mar 3 at 21:20
  • $\begingroup$ Also, how do we know the operation $F/N$ results in exactly $16$ equivalence classes? What is explained in the article is only that the elements of $F$ in $N$ are sent to one equivalence class, and I can’t be sure what happens to the other elements. $\endgroup$ – ensbana Mar 3 at 21:20
  • $\begingroup$ I've added a paragraph for how to do that, which turns out not very complicated as these things go. There is no general method like this that works for arbitrary presentations. $\endgroup$ – Lee Mosher Mar 4 at 1:40
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If you drop $rfrf=1$, you get the free product

$$\Bbb Z_8\ast\Bbb Z_2\cong\langle r,f\mid r^8, f^2\rangle$$

of the presentations $\Bbb Z_8\cong\langle r\mid r^8\rangle$ and $\Bbb Z_2\cong\langle f\mid f^2\rangle$.

Indeed, with a rotation of order two, we get the infinite dihedral group

$$D_\infty\cong \Bbb Z_2\ast\Bbb Z_2\cong\langle r,f\mid r^2, f^2\rangle.$$

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  • $\begingroup$ It appears to me that the article starts with the group $D_8$, then makes the claim that those three relations uniquely determine it from the free group $F$. There’s a gap in between on why it must be those three that I’m not clear about. $\endgroup$ – ensbana Mar 3 at 14:56
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As the other answers have mentioned, if you drop that relation you get the infinite dihedral group. Check out the groups of order 8, and note that conjugation gets written into the relations as $frf^{-1} = r^{-1}$, so $r \mapsto r^{-1}$, for the dihedral group of order $8$, and $frf^{-1} = r$, so $r \mapsto r$ for the direct product $Z_4 \times Z_2$.

This might clear up why you need that particular relation.

And yes, I would agree they are using the first isomorphism theorem as you suggest.

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