# What is the $p$-regular partition corresponding to the sign representation of $S_{n}$ over a field of characteristic $p$?

I'm now interested in the modular representation of symmetric groups.

It is well-known that for a fixed prime $$p$$, there is a bijection between the irreducible representations of $$S_{n}$$ over a field of characteristic $$p$$ and $$p$$-regular partitions of $$n$$.

And it seems that the sign representation of $$S_{n}$$ exists for all odd primes $$p$$.

However, the partition $$(1^{n})$$ of $$n$$ which corresponds to the ordinary (i.e. characteristic 0) sign representation of $$S_{n}$$ is no longer $$p$$-regular if $$p \leq n$$

So I hope to know, which $$p$$-regular partition corresponds to the irreducible 1-dimensional sign representation?

Is there an easy answer?

Thank you very much!

The answer is the $$p$$-regularisation of $$(1^n)$$: this is $$(1^n)$$ if $$n < p$$ and otherwise the partition $$(r^a, (r-1)^{p-1-a})$$ with $$p-1$$ parts, where $$r$$ and $$a \in \{0,1,\ldots, p-1\}$$ are defined uniquely by $$(r-1)(p-1) + a = n,$$ taking $$a = 0$$ if $$p-1$$ divides $$n$$.
For more on this see G. D. James, On the decomposition matrices of the symmetric groups II, J. Algebra 43 (1976), 45--54. The Mullineux Conjecture (now proved) determines the partition labelling a general $$D^\lambda \otimes \mathrm{sgn}$$ for $$\lambda$$ a $$p$$-regular partition. The shortest proof I know is in Example 24.5(iii) in James' Springer lecture notes on the symmetric groups.
Edit. I'd like to record a slightly unusual proof here. By some general theory, the permutation module $$M$$ for a finite group $$G$$ acting on the cosets of $$H \le G$$ defined over a field $$F$$ of prime characteristic $$p$$ has a distinguished Scott module summand. In Green's theory of vertices and sources, this summand has vertex $$Q$$ where $$Q \in \mathrm{Syl}_p(H)$$ and trivial source. It is the unique summand of $$M$$ having the trivial module $$F$$ in its top and socle. In particular, $$P_F$$, the projective cover the trivial module, is the Scott module for trivial vertex.
Now let $$\lambda$$ be a partition of $$n$$ and let $$M^\lambda$$ be the Young permutation module for $$FS_n$$ induced from the Young subgroup $$S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$$. It is known that $$M^\lambda$$ has a distinguished summand $$Y^\lambda$$: this is the unique summand (in any given direct sum decomposition of $$M^\lambda$$) containing the Specht module $$S^\lambda$$. Moreover, $$M^\lambda \cong Y^\lambda \oplus \bigoplus_{\mu > \lambda} c_\mu Y^\mu$$ for some coefficients $$c_\mu$$.
Observe that $$\lambda$$ has a part of size $$p$$ or more if and only if $$S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$$ has a non-trivial Sylow $$p$$-subgroup and so if and only if the Scott module summand of $$M^\lambda$$ is non-projective. Therefore the lexicographically greatest partition $$\mu$$ such that $$M^\mu$$ has $$P_F$$ as a summand is the greatest partition with all parts $$< p$$, namely $$((p-1)^s, t)$$ where $$0 \le t < p-1$$ and $$n = (p-1)s + t$$. Moreover, since $$P_F$$ is not a summand of any $$M^\lambda$$ with $$\lambda > \mu$$ (they all have a Scott module summand, which cannot be project, so is not $$P_F$$), we have $$Y^{((p-1)^s, t)} \cong P_F$$.
This shows that the trivial module appears in the socle of the Specht module $$S^{((p-1)^s,t)}$$, and so using the basic result that $$(S^\lambda)^\star \cong S^{\lambda'} \otimes \mathrm{sgn}$$, the sign module appears at the top of $$S^{((p-1)^s,t))'}$$. Since $$((p-1)^s,t)'$$ is $$p$$-regular, this Specht module has a unique top composition factor and we get $$D^{((p-1)^s, t)'} \cong \mathrm{sgn}$$, as required. (And in agreement with the result from $$p$$-regularisation.)