Representations of A6

In what follows you'll see how to use Axiom to get all the irreducible representations of the alternating group A6 over the field with two elements (GF 2). First, we generate A6 by a three-cycle: x=(1,2,3) and a 5-cycle: y=(2,3,4,5,6). Next we have Axiom calculate the permutation representation over the integers and over GF 2:

Now we apply Parker's 'Meat-Axe' and split it:

We have found the trivial module as a quotient module and a 5-dimensional sub-module. Try to split again:

and we find a 4-dimensional sub-module and the trivial one again. Now we can test if this representation is absolutely irreducible.

and we see that this 4-dimensional representation is absolutely irreducible. So, we have found a second irreducible representation. Now, we construct a representation by reducing an irreducible one of the symmetric group S_6 over the integers mod 2. We take the one labelled by the partition [2,2,1,1] and restrict it to A6:

This gave both a five and a four dimensional representation. Now we take the 4-dimensional one and we shall see that it is absolutely irreducible:

The two 4-dimensional representations are not equivalent:

So we have found a third irreducible representation. Now we construct a new representation using the tensor product and try to split it:

The representation is irreducible, but it may be not absolutely irreducible.

So let's try the same procedure over the field with 4 elements:

Now we find two 8-dimensional representations, dA6d8a and dA6d8b. Both are absolutely irreducible.

and they are not equivalent.

So we have found five absolutely irreducible representations of A6 in characteristic 2. General theory now tells us that there are no more irreducible ones. Here, for future reference are all the absolutely irreducible 2-module representations of A6.

And here again is the irreducible, but not absolutely irreducible representations of A6 over GF 2.