Why are surface groups non-abelian? Let $\Sigma_g=\langle a_1, b_1, \dots,a_n,b_n| \Pi_i [a_i,b_i]\rangle $. I want to show that they are non-abelian for $g\ge2$.
Intuitively I would just look at the commutator $[a_1,b_1]$ and say that the relations ensure that this is non-trivial. But how can I be sure about this? And how would I argue if the considered group has more relations or more complicated ones?
It is easy to see, that the Abelianization is $\mathbb Z^{2n}$. Maybe I could show that $\Sigma_g$ is not isomorphic to it?
 A: First of all, notice that it suffices to show that $\Sigma_2$ is non abelian to show that $\Sigma_g$ is non abelian for $g\geq 2$ : indeed there is a clear surjective morphism $\Sigma_g \to \Sigma_2$ for $g\geq 2$.
Notice also that $\Sigma_1$ is abelian, indeed it's isomorphic to $\mathbb{Z^2}$.
Ok now for $\Sigma_2$, when dealing with groups that are finitely presented there are usually two possibilities to deal with that kind of question : one is a combinatorial proof, e.g. an invariant on reduced words that tells you,say, that $[a_1, b_1]$ is nontrivial; but such a proof can be very hard, especially if you're not used to formal languages and related areas. 
Another solution is to forget the specific presentation with reduced words, and use the universal property of $\Sigma_2$ :  given any group $G$ and any $4$ elements $a,b,c,d$ such that $[a,b][c,d] =1$, there is a unique morphism $\Sigma_2 \to G$ sending $a_1$ to $a$, $b_1$ to $b$ etc.
Thus to prove that $\Sigma_2$ is non abelian, it suffices to find such a $G$ with, say $[a,b]\neq 1$, because then it follows that $[a_1, b_1]\neq 1$.
This type of solution is easier if you're more acquainted with group theory than formal languages for instance. Now there are many many examples of such $G$, here is one :  in $GL_3(\mathbb{F}_2)$, the upper triangular matrix with only $1$'s on the diagonal and a one in the top right corner is a commutator of order $2$: say it's $[a,b]$ (you may find $a,b$ as suitable upper triangular matrices with $1$'s on the diagonal). Then $[a,b][a,b]=1$ but $[a,b]\neq 1$: we have found our group and our elements, and so $\Sigma_2$ is non abelian (even better actually, we know $[a_1,b_1], [a_2,b_2] \neq 1$ !)
There are other examples, for instance every transposition in $\mathfrak{S}\mathbb{N}$ is a commutator of order $2$.
Note : here as $\Sigma_g$ is clearly the fundamental group of an orientable surface of genus $g$, there might be a third solution using topology; though it would probably be painful to write it down properly (and I'm not even sure such a proof is known)
A: The following proof is algebraic. It would be nice to see a topological proof...

You can prove that they are non-abelian (for genus $\geq2$) quite easily with HNN-extensions. To do this, start by writing the presentation as follows:
$$
G=\langle x_1, y_1, x_2, y_2\mid [x_1,y_1][x_2, y_2]\rangle
=\langle x_1, y_1, x_2, y_2\mid x_1^{y_1}x_1^{-1}x_2^{y_2}x_2^{-1}\rangle.
$$
We can take $y_1$ and $y_2$ to be two independent stable letters (note that I am not assuming anything about $G$ here; the map $x_1\mapsto x_1^{-1}$ is always an isomorphism, whether $x_1$ has infinite order or not). Hence, $G$ is a "double" HNN-extension of the group:
$$
H=\langle a, x_1, b, x_2\mid ax_1^{-1}bx_2^{-1}\rangle.
$$
Clearly this group $H$ is free of rank three. By the fundamental result on HNN-extensions, this group $H$ embeds into $G=\pi_1\left(\Sigma_2\right)$. Hence, $\pi_1\left(\Sigma_2\right)$ contains a non-abelian free group as a subgroup, and so is non-abelian as required.
EDIT: actually, you only need to take, for example, $y_2$ as a stable letter. The group $H$ is then $\langle x_1, y_1, x_2, b\mid [x_1, y_1]bx_2^{-1}\rangle$, which is still free of rank three.
A: A proof using your initial idea:
We compare $\mathbb Z^{4n}=\Sigma_{g,ab}$ with $\Sigma_g$.
These groups are not isomorphic. In fact they are not even quasi-isomorphic.
There is a surjective map $\Sigma_g\to F_2=\langle s_1,s_2 \rangle$, given by $a_1 \mapsto s_1$, $a_2\mapsto s_2$.
Thus $\Sigma_g$ has exponential growth while $\mathbb Z^{4n}$ has only polynomial growth.
If you don't want to work with growth functions/ classes of quasi-isometries, you could also try to show by hand that there is no surjection $\mathbb Z^{4n} \to F_2$.
