# 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 $y$.

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I hope your F_{2} is nothing but the field of 2 elements. –  anonymous Oct 18 '10 at 19:43
@Chandru: Of course it is not! It says explicitly that $f$ is a "homomoprhism from the free group on two elements to $G$", so obviously his F_2 was the free group on two elements, not the field with two elements. –  Arturo Magidin Oct 18 '10 at 19:56
@Arturo: Sorry! for the lame mistake! –  anonymous Oct 18 '10 at 20:00

Abelsh, your definition of $f$ does not make sense: you cannot define the image of $x$ to be $(12)(34)$ unless you already know that $G$ is a group of permutations.

Instead, what you want to do is define $f$ from $\mathbf{F}_2$ to $A_4$ by defining it as you want: $f(x) = (12)(34)$ and $f(y)=(123)$. If you do this, then it is straightfoward to check that the images satisfy the given relations, $(f(x))^2 = 1$, $(f(y))^3 = 1$, and since $f(x)f(y) = (12)(34)(123) = (243)$ (I'm composing my permutations right to left), then $(f(x)f(y))^3 = 1$. That means that the kernel of $f$ contains $x^2$, $y^3$, and $(xy)^3$, and therefore, by van Dyck's Theorem, the homomorphism factors through $G$. That means that $G$ has $A_4$ as a quotient, and thus has at least 12 elements.

But unless you find a precise description of the kernel, it would be difficult to prove that the kernel is exactly the subgroup generated by $x^2$, $y^3$, and $(xy)^3$. Instead, you might want to find a normal form for elements of $G$ to show that $G$ has at most 12 elements; if you can do that, you get that $G$ has exactly $12$ elements, and has $A_4$ as a quotient, which gives you what you want.

So... do we have a sufficiently nice normal form for elements of $G$? Every element of $F$ can be written as a word in $x$ and $y$; using the relations in $G$ we can replace any power of $x$ with either $x^0$ or $x$, and any power of $y$ with either $y^0$, $y^1$, or $y^2$. So every element of $G$ can be written as a string that alternates $x$s and either $y$'s or $y^2$'s. And $(xy)^3 = 1$ so $(y^2x)^3=1$. That means that you can replace any string of the form $xyx$ with $y^2xy^2$, and replace $xy^2x$ with $yxy$. Note that in each of these replacements, you go from a string with two $x's$ to a string with just one $x$. That means that you can take any string that alternates $xs$ with $y$s or $y^2$s and use those replacements to get it down to one that has at most one $x$ in it. How many such words are there? That gives an upper bound for the number of distinct elements that $G$ can have.

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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.

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You are screaming for inclusion here mathoverflow.net/questions/42512/… :) –  Mariano Suárez-Alvarez Oct 19 '10 at 4:40
Mariano, of course I see what you mean, but this is genuinely how I think about this. It's not so much about machinery but language. There's certainly an elementary proof buried in here somewhere. –  HJRW Oct 19 '10 at 4:52