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Note: All Young diagrams are to use the English notation scheme.

Suppose I have two tableaux $T$ and $T'$ on the same Young diagram (we insert the numbers $1, 2, \ldots, n$ in two different ways in the diagram). We say that $T > T'$ if the first entry (while reading left to right, top to bottom) in which $T$ and $T'$ differs is such that the entry of $T$ is larger than that entry of $T'$.

Is it true that if $T > T'$, then there exist $a \neq b$, $\{a, b\} \subset \{1, 2, \ldots, n\}$ such that $\{a, b\}$ appears in the same row of $T$ and $\{a, b\}$ appears in the same column of $T'$?

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    If you are going to say «left to right, top to bottom» you need explain how you draw the diagrams, for there are *two* different conventions (the English and the French conventions...) which will result in two different orders.2012-02-23
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    Woah. Did not realize that there were English/French conventions. Thanks, edited.2012-02-23
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    I suspect you mean *standard* tableaux? If not, you might as well write $T\ne T'$, since the order wouldn't matter; also the statement would be evidently false, since you could have different tableaux on a diagram with only one row or only one column.2012-02-23

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Suppose $1,2,3,\dots,17$ appear in the same places in $T$ and $T'$, but $17+1=18$ is in location $X$ in $T$ and location $Y$ in $T'$, with $X\ne Y$. Convince yourself that $X$ and $Y$ can't be in the same row, and can't be in the same column. Assume without loss of generality that $X$ is above and to the right of $Y$ (I don't know if this is English or French; my entries increase left to right, and top to bottom). Consider the location at the intersection of the row containing $X$ and the column containing $Y$. The number in that location is in the same row as 18 in $T$ and in the same column as 18 in $T'$.

Of course, 17 is a variable.

EDIT: Let's look at an example. Suppose your two tableaux both look like $$\matrix{1&2&5&6&12&13&a\cr3&7&8&10&17&b&\cr4&14&15&c&&&\cr9&16&d&&&&\cr11&e&&&&&\cr f&&&&&&\cr}$$ That is, they agree up to 17. In each, 18 must go to one of the six locations $a,b,c,d,e,f$. Suppose it goes to $b$ in $T$ and to $d$ in $T'$. The row containing $b$ and the column containg $d$ meet at $8$, so $8$ and $18$ are in the same row in $T$ and in the same column in $T'$. Similarly for any other choice of two of the 6 possible locations for $18$.

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    Are you sure this works for non-prime values of $17$?!2012-02-24
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    I'm sure you're kidding, but just for the record the divisibility status of 17 was not used anywhere in the proof.2012-02-24
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    This is the English style.2012-02-25