The order deduced from relations in $D_n$ If $D_n \triangleq \langle a,b | a^n=e, b^2=e, abab=e \rangle$, can it be proved that the order of $a, b$ is actually $n$ and $2$ respectively ?
I mean can the relations on the right somehow after some arithmetics produce something like $a^{n-1}=e$? It seems unlikely but how to prove it?
 A: Actually proving anything about a group presented using generators and relations involves tricky combinatorial maneuvers -- unless you use the universal property. Usually "proofs" not using the universal property appeal to intuition and are incomplete and kind of hand-wavy. Here's a "solid" proof.
The universal property: 
Let $G=\langle X|R \rangle$ ($G$ is generated by the elements of $X$ with relations $R$). 
Then let $f$ be any function from $X$ to some group: $f:X \to H$ 
where given any relation: $x_1^{e_1}x_2^{e_2}\cdots x_n^{e_n} \in R$ (here $x_i \in X$ and $e_i = \pm 1$) we have $f(x_1)^{e_1}f(x_2)^{e_2}\cdots f(x_n)^{e_n}=1$ (essentially images of relations must map to the identity).
Then there exists a unique homomorphism $\hat{f}:G \to H$ such that $\hat{f}(x) = f(x)$ for all $x \in X$.
This universal property is really the definition of what we mean by a group presented via generators and relations.
So $D_n = \langle a,b \;|\; a^n=1,b^2=1,abab=1 \rangle$ has such a universal property.
Consider $f:\{a,b\} \to H$ defined by $f(a)=R_{360^\circ/n}$ and $f(b)=V$ where $H$ is the group of isometries of a regular $n$-gon, $R_{360^\circ/n}$ is a rotation of $360^\circ/n$, and $V$ is some reflection in $H$ ($H$ is a concrete realization of $D_n$). Notice that the relation $a^n=1$ translates to $f(a)^n=R_{360^\circ/n}^n=R_{0^\circ}$, $b^2=1$ translates to $f(b)^2=V^2 =R_{0^\circ}$. Finally, $abab=1$ translates to $f(a)f(b)f(a)f(b)=R_{360^\circ/n}VR_{360^\circ/n}V=WW$ where $W$ is some reflection so $f(a)f(b)f(a)f(b)=R_{0^\circ}$. Thus all relations are satisfied.
Therefore, there exists a unique homomorphism extending $f$, call it $\hat{f}:D_n\to H$. 
Now, $\hat{f}(a)=f(a)=R_{360^\circ/n}$ so the order of $\hat{f}(a)$ is $n$. Thus the order of $a$ must be a multiple of $n$. However, $a^n=1$ so the order of  $a$ is no more than $n$. Thus it is $n$.
We could have used $2 \times 2$ matrices to realize $D_n$ (but that's too much trouble to typeset). Also, without too much more work, we could have proved that $\hat{f}$ is actually an isomorphism.
To prove the order of $b$ is 2 we can use the concrete group $\mathbb{Z}_2$. Define $f:\{a,b\}\to \mathbb{Z}_2$ by $f(a)=0$, $f(b)=1$ (in $\mathbb{Z}_2$). Let's verify the relations (keep in mind that we use additive notation for $\mathbb{Z}_2$): $nf(a)=n0=0$, $2f(b)=2\cdot 1=0$, $f(a)+f(b)+f(a)+f(b)=0+1+0+1=0$. 
Therefore, there exists a unique homomorphism extending $f$, say $\hat{f}:D_n \to \mathbb{Z}_2$. Again, $\hat{f}(b)=1$ has order 2, so $b$'s order must be a multiple of 2. However, $b^2=1$ so its order is no more than 2. Therefore, its order is exactly 2.
In the end, you prove things about your generator/relation group by using concrete groups (concrete homomorphic images).
Note: It is tempting to try to prove the order of $a$ is $n$ using a group like $\mathbb{Z}_n$, but $\mathbb{Z}_n$ isn't a homomorphic image of $D_n$ (in general) so this doesn't work. :(
