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I try to improve my understanding of the dihedral group. One way of presentation of the dihedral group $D_n$ of order $2n$ is $$\langle a,b : a^2=b^2=(ab)^n=1 \rangle.$$

After a moment of thought it seemed pretty 'obvious' to me that the set of all group elements could be written as $G= \lbrace (ab)^k, (ab)^ka:k=0,...,n-1 \rbrace$. It was easy to show that G is a group. Unfortunately I could not prove that the set $G$ indeed represents the full group $D_n$.

Mor precisely I have trouble to show that all elements in the above set $G$ are pairwise different and that there are no other elements of $D_n$ not contained in G.

E.g. why is it not possible that $(ab)^k=1$ for some $k=1,...,n-1$?

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  • $\begingroup$ Typically one would use the universal property of quotients, and a concrete description of the dihedral group(or whatever group you are working with). You can show that $D_n$ has a generating set satisfying those relation, and that means $D_n$ is a quotient of the presentation, so the presentation has at least $2n$ elements. You already showed that the group has at most $2n$ elements so you know have a bijection. $\endgroup$
    – user29123
    Dec 21, 2015 at 15:30

2 Answers 2

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Recall that when we say that $G = \langle a, b \, | \, a^2 = b^2 = (ab)^n = 1\rangle$, what we mean is that $G$ is the quotient of the free group $\langle a, b\rangle$ by the normal subgroup $N$ generated by $a^2, b^2, (ab)^n$. Now, let's concretely view $D_n$ as the group of rotations and reflections of the regular $n$-gon which preserve the vertices. I'll assume you're familiar with this group.

We can define a group homomorphism $\varphi :\langle a,b\rangle \to D_n$ by sending $a$ and $b$ to "adjacent" reflections. By this, I simply mean that $\varphi(ab) = \varphi(a)\varphi(b)$ should be a rotation of order $n$. Using our knowledge of $D_n$, it's easy to confirm that $a^2, b^2$ and $(ab)^n$ are in the kernel of $\varphi$. Therefore all of $N$ is contained in the kernel. It follows that there is an induced group homomorphism $$\overline{\varphi}: \langle a, b \, | \, a^2 = b^2 = (ab)^n = 1\rangle \to D_n$$ by the univeral property of the quotient. Moreover, you have shown that the domain of the map has at most $2n$ elements, and by construction $\overline{\varphi}$ is surjective (since $\varphi$ is). Since $|D_n| = 2n$ as well, $\overline{\varphi}$ must be bijective, so we're done.

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  • $\begingroup$ To be more precise, we need to prove that $N\trianglelefteq F_2=\langle a,b\rangle$. The map $\psi:F_2/\ker \varphi\to D_n$ is bijective, since $\varphi$ is surjective. Hence, $2n=[F_2:\ker \varphi]=\frac{|G|}{[\ker\varphi:N]}\le \frac{2n}{[\ker\varphi:N]}$, so $\ker\varphi=N$. $\endgroup$ May 14, 2020 at 2:06
  • $\begingroup$ @hlcrypto123 $N \trianglelefteq F_2$ by construction (it's the normal subgroup generated by $a^2, b^2, (ab)^n$) $\endgroup$
    – Alex G.
    May 16, 2020 at 21:33
  • $\begingroup$ I see. So $N$ need not be the same as the subgroup generated by those 3 elements. Thus, $N$ is the intersection of all normal subgroups of $F_2$ containing those elements, amongst one of which, is $\ker\varphi$. $\endgroup$ Jun 7, 2020 at 20:24
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A concrete way of showing that a lower power of $ab$ could not be the identity is by constructing an action of the group by symmetries of a regular $n$-gon. Then a lower power of $ab$ will correspond to a nontrivial rotation, so certainly could not be the identity of the group.

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