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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 $m=3$.

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  • $\begingroup$ How do you define the rank modulo composite number? $\endgroup$ Nov 28 '21 at 11:00
  • $\begingroup$ @FedorPetrov It is the rank as a matrix with integer entries I think. You just take modulo to obtain this integer matrix. That is what GAP does when it comuputes the rank of a matrix modulo $m$ (I think). $\endgroup$
    – Mare
    Nov 28 '21 at 11:01
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    $\begingroup$ so, say, for $m=2$ it is not the same as the rank over $\mathbb{F}_2$? $\endgroup$ Nov 28 '21 at 11:07
  • $\begingroup$ @FedorPetrov It is probably not in general. I edited what I mean with the matrix modulo $m$ (it is just the integer matrix where each entry is replaced by its canonical representatitive modulo m, for example -1 gets replaced by m-1). $\endgroup$
    – Mare
    Nov 28 '21 at 11:13
  • $\begingroup$ Please could you check the sequence of ranks for $m=4$: the $2 \times 2$ character table of $S_2$ cannot have rank $4$. It looks like you omitted the initial $2$, since I make the sequence $2, 3, 4, 6, 9, 12, 16, 22, 29, 38, 50, \ldots$. Your conjecture is true when $m=4$ for $n \le 15$. $\endgroup$ Nov 28 '21 at 11:45
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This is true when $m$ is prime and false in general.

Counterexample. Take $S_8$ with $m=6$. Computer calculations show that the $\mathbb{Z}$-rank of the character table of $S_8$ with entries taken modulo $6$ to lie in $\{0,1,2,3,4,5\}$ is $22$. Thus this matrix has full rank, equal to its number of columns. (Its rank as a matrix with entries in $\mathbb{Z}/6\mathbb{Z}$ is $13$.) Since $(6,2)$ and $(6,1,1)$ are the partitions of $8$ with a part divisible by $6$, there are exactly $20$ partitions of $8$ into parts not divisible by $6$.

Proof when $m$ is prime. Let $\chi^\lambda$ be the irreducible character of $S_n$ canonically labelled by the partition $\lambda$. Let $\phi^\mu$ be the $p$-modular Brauer character of $S_n$ labelled by the $p$-regular partition $\mu$. (A partition is $p$-regular if it has at most $p-1$ parts of any given size: such partitions label the simple representations of $S_n$ in characteristic $p$.) There exist decomposition numbers $d_{\lambda\mu} \in \mathbb{N}_0$ such that

$$\chi^\lambda(g) = \sum_{\mu} d_{\lambda\mu} \phi^\mu (g)$$

for all $p$-regular $g \in S_n$. (Here $p$-regular means 'of order not divisible by $p$'.) Hence the span of the columns of the character table labelled by $p$-regular partitions is the same as the span of the columns of the Brauer character table. Since the number of $p$-regular partitions is the number of $p$-regular conjugacy classes (the Brauer character table is square), the rank of this column submatrix is the number of partitions of $n$ into parts coprime to $p$.

Finally the span of the columns of the character table labelled by $p$-singular elements, taken modulo $p$, is contained in the span of the columns labelled by $p$-regular elements, since if $g = g_{p'}h$ is the factorization of $g \in S_n$ into its $p$-regular part $g_{p'}$ and a $p$-element $h$ then $\chi(g) \equiv \chi(g_{p'})$ mod $p$ for any character $\chi$.

Reference for character congruence. The last fact is (12) in this paper of R. Brauer where it's called 'a well known congruence'. It can be seen in Mare's example: the fourth column is for the conjugacy class of non $3$-regular elements with cycle type $(3,1,1)$, i.e. $3$-cycles, and is equal modulo $3$ to the column for the identity element.

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    $\begingroup$ Great answer, thanks. This might raise the question, whether it is true when m is a power of a prime (or for which $m$ it is true at all). I will do some tests. $\endgroup$
    – Mare
    Nov 28 '21 at 12:07
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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 $m$.

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