= Solution
Write the <transposition> as $\tau=rc$ with $r\in R(t)$ and $c\in C(t)$. If $x$ is not one of the two entries moved by $\tau$, then $r(c(x))=x$. The entries $x$ and $c(x)$ lie in one column of $t$, while $c(x)$ and $r(c(x))=x$ lie in one row. A <Young diagram> has only one cell at the intersection of a specified row and column, so $c(x)=x$ and then $r(x)=x$.
Consequently both $r$ and $c$ fix every entry outside the support of $\tau$. On the two remaining entries each is either the identity or their transposition. They cannot both transpose them, since then $rc=1$, and they cannot both be the identity. Exactly one of $r,c$ is therefore $\tau$, proving that $\tau$ lies in exactly one of $R(t)$ and $C(t)$.
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