Understand the free group universal property applied to $D_n$ 
For $n ≥ 3$ and $D_n$ the dihedral group of order $2n$ has the presentation
$$\langle r, s : r^n = s^2 = srsr = 1\rangle.$$
Prove that for all $(a, b) \in (\Bbb Z/n\Bbb Z)^2$, there exists a morphism  $f$ verifying $f(r) = r^a , f (s) = r^b s$.

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.
 A: You're asked to show that for every $(a,b) \in (\mathbb{Z}/n)^2$, there is a group hom 
$f_{(a,b)} : D_{n} \to D_{n}$
defined by


*

*$f_{(a,b)}(r) = r^a$

*$f_{(a,b)}(s) = r^b s$

Alright, how can we do this?
The given presentation $D_n = \langle r,s ~|~ r^n = s^2 = srsr = 1 \rangle$ says
$D_n$ is a quotient of $\langle r, s \rangle$ (the free group with generators $r$ and $s$), and the kernel $N$ is generated by the relations shown.
Using the universal property of the free group, we know $f_{(a,b)}$ as defined above is a perfectly good function from $\langle r, s \rangle \to D_n$. We said what it does to generators, easy.
But we don't care about free groups, we care about $D_n$!
To show that $f_{(a,b)} : \langle r, s \rangle \to D_n$ descends to a map 
$D_n \to D_n$, we use the universal property of the quotient group. 
Since $D_n = \langle r, s \rangle / N$, all we need to do is show that $N \subseteq Ker f_{(a,b)}$.
The last piece of the puzzle is to remember $N$ is the smallest normal subgroup such that $r^n, s^2, srsr \in N$. So any normal subgroup $K$ containing $r^n$, $s^2$, and $srsr$ must also contain $N$.
Thus, it suffices to show that $r^n, s^2, srsr \in Ker f_{(a,b)}$.


*

*$f_{(a,b)}(r^n) = (r^a)^n = (r^n)^a = 1^a = 1$

*$f_{(a,b)}(s^2) = (r^b s)(r^b s) = (r^b)(s r^b s) = r^b r^{-b} = 1$

*$f_{(a,b)}(srsr) = (r^b s)(r^a)(r^b s)(r^a) = s r^{-b} r^a r^b r^{-a} s = s1s = 1$
So we see $N \subseteq Ker f_{(a,b)}$ for each $(a,b)$. The claim follows.

I hope this helped ^_^
