The results we showed last time have a few immediate consequences we will have use of.
To see this, let and be two entries in the same row of . They cannot be in the same column of , since if they were then the swap would be in the column-stabilizer . Then we could conclude that , which we assumed not to be the case. But if no two entries from the same row of are in the same column of , the dominance lemma tells us that .
Now if it turns out that it’s not surprising that . Luckily in that situation we can say something interesting:
Indeed, we must have for some , basically by the same reasoning that led to the dominance lemma in the first place. Indeed, the thing that would obstruct finding such a is having two entries in some column of needing to go on the same row of , which we know doesn’t happen. And so we calculate
Now if is any vector in the Specht module, and if is a tableau of shape , then is some multiple of . Indeed, we can write
were the are -tableaux. For each one of these, we have . Thus we find
which is a multiple of , as asserted.