Again, we pick some canonical Young tableau of shape so that every other tableau can be written uniquely as for some . That is, the set of all Young tabloids is the orbit of the canonical one. By general properties of group actions we know that there is a bijection between the orbit and the index of the stabilizer of in . That is, we must count the number of permutations with row-equivalent to .
It doesn’t really matter which we pick; any two tableaux in the same orbit — and they’re all in the same single orbit — have isomorphic stabilizers. But like we mentioned last time the usual choice lists the numbers from to on the first row, from to on the second row, and so on. We write for the stabilizer of this choice, and this is the subgroup of we will use. Notice that this is exactly the same subgroup we described earlier.
Anyway, now we know that Young tabloids correspond to cosets of ; if for some , then
So we can count these cosets in the usual way:
How big is ? Well, we know that
Since it will come up so often, we will write this product of factorials as for short. We can then write and thus we calculate for the number of cosets of in . And so this is also the number of Young tabloids of shape , and also the dimension of .
Now, along the way we saw that the Young tabloid corresponds to the coset . It should be clear that the action of on the Young tabloids is exactly the same as the coset action corresponding to . And thus the permutation module must be isomorphic to the induced representation .