# Transitive action of column stabilizers on tabloids

In the representation theory of the symmetric group $$S_n$$, important concepts are partitions $$\lambda$$ of $$n$$ and $$\lambda$$-tableaux $$t$$. The symmetric group acts on $$\lambda$$-tableaux in the obvious way, and gives a transitive action.

Two $$\lambda$$-tableaux $$t$$ and $$s$$ are said to be row equivalent if they have the same rows -- the resulting equivalence classes are called $$\lambda$$-tabloids and the equivalence class of $$t$$ denoted $$\{t\}$$.

The action of $$S_n$$ on $$\lambda$$-tableaux is compatible with row equivalence and so descends to an action on the set of $$\lambda$$-tabloids. Of course the action is still transitive.

Now each tableau $$t$$ determines its column stabiliser $$C_t \leq S_n$$ - the subgroup of permutations not changing the columns of $$t$$.

In understanding that the Specht modules are irreducible representations, a key point seems to be that for each $$\lambda$$-tableau $$t$$ and $$\lambda$$-tabloid $$s$$ there exists $$\pi \in C_t$$ such that $$\pi \{t\} = \{s\}$$. (This is clearly equivalent to the statement that for each tableau $$t$$ the action of $$C_t$$ on the set of $$\lambda$$-tabloids is transitive.)

This is claimed (without further demonstration) in Lemma 4.6 of James' book - Representations of the symmetric group, and in the corresponding result Corollary 2.4.2 of Sagan's book - The symmetric group. I believe, therefore, that it must not be very hard, but I don't immediately see why it is true - can anyone offer some help / a proof?

Thanks in advance.