TITLE: The character table of the symmetric group modulo m
QUESTION [11 upvotes]: Let $S_n$ be the symmetric group and $M_n$ the character table of $S_n$ as a matrix (in some order) for $n \geq 2$.

Question: Is it true that the rank of $M_n$ as a matrix modulo $m$ for $m \geq 2$ is equal to the number of partitions of $n$ by numbers that are not divisible by $m$?

Here the matrix modulo $m$ is obtained by replacing each number $l$ by its canonical representative mod $m$ (for example -1 mod 3 =2) and then calculate the rank of the obtained matrix as a matrix with integer entries.
This seems to be true for $m=2,3,4$ by some computer experiments.
For $m=2$ the sequence of ranks (for $n \geq 2$) starts with 2,3,4,5,6,8,10,12,15,18,22 , for $m=3$ it starts with 2,4,5,7,9,13,16,22 and for $m=4$ it starts with 2,3,4,6,9,12,16,22,29.
For example for $n=5$ and $m=3$, the matrix $M_5$ looks as follows:
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & -1 & 1 & 1 & -1 \\
2 & 0 & 2 & -1 & 0 \\
3 & -1 & -1 & 0 & 1 \\
3 & 1 & -1 & 0 & -1 
\end{bmatrix}
modulo 3 the matrix is given by
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 1 & 1 & 2 \\
2 & 0 & 2 & 2 & 0 \\
0 & 2 & 2 & 0 & 1 \\
0 & 1 & 2 & 0 & 2 
\end{bmatrix}
and this matrix has rank 4.
The paritions of $n=5$ are given by [ [ 1, 1, 1, 1 ], [ 2, 1, 1 ], [ 2, 2 ], [ 3, 1 ], [ 4 ] ] and thus there are 4 partitions of 5 whose parts are not divisble by $m=3$.

REPLY [21 votes]: When $m$ is prime there is a simpler proof. The Smith normal form of the character table of $S_n$ is computed at Problem 14  here (solution here). From this it follows that the rank of the character table mod  $m$ equals the number of partitions of $n$ for which every part has multiplicity less than $m$. By a simple generating function argument (generalizing Euler's well-known proof for the case $m=2$) this is also the number of partitions of $n$ for which no part is divisible by $m$.