Generators and Relations for $A_4$ Let $G=\{x,y|x^2=y^3=(xy)^3=1\}$
I would like to show that $G$ is isomorphic to $A_4.$
Let $f:\mathbf{F}_{2} \to G$ be a surjective homomorphism from the free group on two elements to $G$. Let $f$ map $x \to (12)(34)$ and $y \mapsto (123)$. I'm not sure how to show that these elements generate the kernel of $f$. If they do generate the kernel, how do I conclude that the order of $G$ is $12?$
Once I have that the order of the group is of order 12 then I can show that $G$ contains $V$ (the Klein four group) as a subgroup, or that $A_4$ is generated by the image of $x$ and $y$.

Perhaps this answer will use too much technology.  Still, I think it's pretty.
Consider $A_4$ as the group of orientation-preserving symmetries of a tetrahedron $S$.  The quotient $X=S/A_4$ is a 2-dimensional orbifold.  Let's try to analyse it.
Two-dimensional orbifolds have three different sorts of singularities that set them apart from surfaces: cone points, reflector lines and corners where reflector lines meet.  Because $A_4$ acts preserving orientation, all the singularities of $X$ are cone points, and we can write them down: they're precisely the images of the points of $S$ that are fixed by non-trivial elements of $A_4$, and to give $X$ its orbifold structure you just have to label them with their stabilisers.
So what are these points?  There are the vertices of $S$, which are fixed by a rotation of order 3; there are the midpoints of the edges of $S$, which are fixed by a rotation of order 2; and finally, the midpoints of faces, which are fixed by a rotation of order 3.
A fundamental domain for the action of $A_4$ is given by a third of one of the faces, and if you're careful about which sides get identified you can check that $X$ is a sphere with three cone points, one labelled with the cyclic group $C_2$ and the other two labelled with the cyclic group $C_3$.
Finally, we can compute a presentation for $A_4$ by thinking of it as the orbifold fundamental group of $X$ and applying van Kampen's Theorem.  This works just as well for orbifolds, as long as you remember to consider each cone point as a space with fundamental group equal to its label.
The complement of the cone points is a 3-punctured sphere, whose fundamental group is free on $x,y$.  The boundary loops correspond to the elements $x$, $y$ and $xy$.  Next, we take account of each cone point labelled $C_n$ by inserting a relation that makes the $n$th power of the appropriate boundary loop equal to $1$.  So we get the presentation
$\langle x,y\mid x^2=y^3=(xy)^3=1\rangle$
as required.
