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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.

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