## Specht Modules

Till now, our description of the irreps of are rather abstract. It would be helpful to have a more concrete construction of these representations – one way is via Specht modules. First write

Thus if , the only common irrep between and is This suggests that we should look at those two representations in more concrete terms.

## General Result

Lemma. Let be a finite group acting transitively on a non-empty finite set X. Pick and let be its stabilizer group. Then for any one-dimensional character ,as left -modules, where

**Proof**

First note that if , then since:

On the other hand, if , is not a multiple of since the coefficient of *e* is 0.

Now define the map as follows: for take it to The map is well-defined and injective since:

Surjectivity is clear; it is also *G*-equivariant since:

Hence we get an isomorphism of left -modules. ♦

## Young Symmetrizer

In particular, consider the case where and let be either the trivial or the alternating character Given a partition , fix a filling of the Young diagram of with such that each element occurs exactly once. E.g. for let us pick:

Now let (resp. ) be the set of all which takes every element in the *i*-th row (resp. column) to an element of the *i*-th row (resp. column). Thus, in our above example:

where is the group of all permutations of set *A*. By the above general lemma, we have:

Now let us define a common component of and

Definition. TheYoung symmetrizerof the filling is defined by:

Note that is contained in both and since

Hence is either 0 or It has to be the latter since the coefficient of *e* in is 1 (because ). Similarly, we see that is also isomorphic to

## Example

Suppose *d*=3 and . Pick the filling *T* by placing 1, 2 at the top row, then 3 below. We have *R*(*T*) = {*e*, (1,2)} and *C*(*T*) = {*e*, (1,3)}. This gives:

The module has a basis where