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For $n ≥ 3$ and $D_n$ the dihedral group of order $2n$ with présentation $\langle r, s : r^n = s^2 = srsr = 1\rangle$

prove that for all $(a, b) \in (\Bbb Z/n\Bbb Z)^2$, there exists a morphism $f$ vérifying $f(r) = r^a , f (s) = r^b s$.

in the "hint" solution I have, it mentions the following:

The universal properties of free groups and the quotient show that $f$ is well defined if $r^a$ and $r^b$ s verify the relations of $r$ and $s$.

I don't understand, what is exactly meant by the free group? is it $\Bbb Z$ or $D_{2n}$. I know what is a free group but I can't really connect the dots.

Thanks for any help.

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    $\begingroup$ The morphism $f$ is from where to where? $\endgroup$ – giannispapav Apr 15 at 17:15
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    $\begingroup$ The free group is the free group on the letters $r$ and $s$. $D_n$ is the quotient of this free group by the relations $r^n, s^2,$ and $srsr$. $\endgroup$ – jgon Apr 15 at 17:23
  • $\begingroup$ It is not stated clearly because I've written the question litteraly from my exam and I guess $f$ is $D_n \to D_n$ where $n$ is the Dihedral group of order $2n$ $\endgroup$ – PerelMan Apr 15 at 17:28

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