# The character table of the symmetric group modulo m

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 $$m=3$$.

• How do you define the rank modulo composite number? Nov 28 '21 at 11:00
• @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).
– Mare
Nov 28 '21 at 11:01
• so, say, for $m=2$ it is not the same as the rank over $\mathbb{F}_2$? Nov 28 '21 at 11:07
• @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).
– Mare
Nov 28 '21 at 11:13
• 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 $n \le 15$. Nov 28 '21 at 11:45

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.

• 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.
– Mare
Nov 28 '21 at 12:07

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 $$m$$.