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I know that a dihedral group $D_{2n }$ is generated by rotation $r$ by $2\pi/n$ and reflection $s$ subject to relations $r^n = 1$, $s^2 = 1$, and $rs = sr^{-1}$. So a dihedral group $D_{2n}$ has a presentation $(r, s {\,|\,} r^n, s^2, (rs)^2)$. In other words, $$ D_{2n} \cong (r, s {\,|\,} r^n, s^2, (rs)^2) = F(r,s)/N $$ where $N$ is the smallest normal subgroup containing $\{r^n, s^2, (rs)^2\} \subset F(r, s)$. It is not difficult to show, without using the isomorphism, that there are at most $2n$ equivalence classes in $F(r,s)/N$, namely, $[r^is^j]$ for $i = 0, \dots, n-1$ and $j = 0, 1$. Then the isomorphism implies that all of them are distinct. Is it possible to show that these equivalence classes are all distinct without using the isomorphism? For example, how to show that $r \nsim ()$, or, equivalently, that $r \notin N$?

Edit: I don't think the link has an answer to my question. So let me rephrase it. Consider a free group $F(a,b)$. Let $N$ be the smallest normal subgroup that contains "words" $a^n$, $b^2$, and $(ab)^2$. Is it possible to show that $a \notin N$ without using any specific group "from nature"?

Yerbolat
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    The general method would be to find a homomorphism to a group "in nature" which has $2n$ elements. The linked answer does exactly that. – Lee Mosher Jun 12 '22 at 22:04
  • Is it possible to do that without using a specific group "from nature"? @Lee Mosher – Yerbolat Jun 12 '22 at 22:12
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    It might be possible, but I'm not aware of a way to do it. Here's some intuition from mathematical logic: The rewrite rules for $G$ give us a proof system which lets us verify that two words represent the same group element. Indeed, a "proof" of this fact is just a sequence of applications of the relations which converts one word into another. What you want to do is argue that two words are different, that is, you want to show that there is no proof that they are the same. – Chris Grossack Jun 12 '22 at 23:21
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    But generally, there's no way to argue syntactically that no proof exists! The way we show that no proof exists is by building a model which verifies the axioms, but in which the conclusion fails. In this group theoretic setting, that means to show that two elements are different, our only recourse is to find a homomorphism from our group $G$ to some other group $H$ so that our two words get sent to different elements of $H$! Of course, this is exactly the approach that @LeeMosher is suggesting. If a purely syntactic approach exists, I would love to see it, but I suspect there isn't one. – Chris Grossack Jun 12 '22 at 23:24
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    @HallaSurvivor Interesting! Thank you for sharing insights. I do understand that it is possible to do it constructing a homomorphism. Just was extremely curious if it is possible to argue "syntactically". – Yerbolat Jun 12 '22 at 23:29

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