Higher Congruences in Character Tables

We proceed by column orthogonality of the unramified character table. Let $C_{\Gamma}(g)$ be the $\Gamma$-conjugacy class of $g\in G$, and let $\delta_{C_{\Gamma}(g)}$ be the $\Gamma$-class function

By Proposition 2.3, the unramified irreducible characters form a basis for $\Gamma$-class functions, so $\delta_{C_{\Gamma}(g)}=\sum_{\chi}\langle(\delta_{C_{\Gamma}(g)},\chi\rangle/\langle\chi,\chi\rangle)\chi$, where the sum is over unramified irreducible characters. By definition $\langle\delta_{C_{\Gamma}(g)},\chi\rangle=|C_{\Gamma}(g)|\chi(g^{-1})/|G|$. Thus for $g_{1},g_{2}\in G$,

Suppose $g_{1},g_{2}\in G$ are not $\Gamma$-conjugate. Let $M$ be the least common multiple of $\langle\chi,\chi\rangle$ over the unramified irreducible characters $\chi$. Then $\sum_{\chi}\chi(g_{1})\chi(g_{2}^{-1})/\langle\chi,\chi\rangle=0$, while

The left-hand side is an integer combination of unramified character values, while the right-hand side has $p$-adic valuation at most $\operatorname{ord}_{p}M+\operatorname{ord}_{p}|G|=e+f$. If $\chi(g_{1})\equiv\chi(g_{2})\mod p^{e+f+1}$ for all $\chi$, then we obtain

a contradiction. Hence $g_{1}$ and $g_{2}$ are $\Gamma$-conjugate, as desired. ∎

If $G$ is a $p$-group, then $g_{1}$ and $g_{2}$ are $\Gamma$-conjugate if and only if $g_{1}$ and $g_{2}$ generate conjugate cyclic subgroups. So let $C$ be the cyclic subgroup generated by $g_{1}$. Consider $\chi_{1}=\operatorname{Ind}_{C}^{G}1$. Then $\chi_{1}(g_{1})=[N_{G}(C):C]$, so $\operatorname{ord}_{p}(\chi_{1}(g_{1}))\leq e$, while $\chi_{1}(g)=0$ if $g$ is not conjugate to an element of $C$. As $\chi_{1}(g_{2})\equiv\chi_{1}(g_{1})\neq 0\mod p^{e+1}$, $\chi_{1}(g_{2})$ is not zero. Hence, $g_{2}$ is conjugate into $C$.

Now let $\chi_{2}=\operatorname{Ind}_{C^{p}}^{G}1$. The same argument shows $\operatorname{ord}_{p}(\chi_{2}(g_{2}))\leq e$ if $g_{2}$ is conjugate into $C^{p}$, while $\chi_{2}(g_{1})=0$. Hence $g_{2}$ is not conjugate to an element of $C^{p}$. Since $C$ is a $p$-group, this implies $g_{2}$ is conjugate to a generator of $C$, so $g_{1}$ and $g_{2}$ generate conjugate cyclic subgroups of $G$. ∎

Since we are taking a trace, we may replace $V$ with its semisimplification under the action of $K[A,B]$. By the Nullstellensatz, the simple $K[A,B]$-modules are of the form $L$ for $L/K$ a finite extension generated by $A,B\in L$. Since $A$ and $B$ have order a power of $p$, $L=K(\zeta)$ for some primitive $p^{j}$th root of unity $\zeta$, and $A$ and $B$ are both powers of $\zeta$. Since $p$ is unramified in $K$, the trace form for $\xi\in\langle\zeta\rangle$ is given by

so that $tr_{L/K}=\chi_{1}-\chi_{2}$ for $\chi_{1}$ the regular character of $\langle\zeta\rangle$ and $\chi_{2}$ the regular character of $\langle\zeta\rangle/\langle\zeta^{p^{j-1}}\rangle$.

Thus, it suffices to show that if $C$ is a cyclic $p$-group with regular character $\chi$ and $A,B\in C$, then $\chi(AB^{p^{m-1}})\equiv\chi(A)\mod p^{m}$. The only case when $\chi(AB^{p^{m-1}})\neq\chi(A)$ is if exactly one of $AB^{p^{m-1}}$ and $A$ is not the identity. Hence $B^{p^{m-1}}$ is not the identity, so $C$ must have order greater than $p^{m-1}$. Thus $p^{m}\mid|C|$. But the nonzero value of $\chi$ is $|C|$. Hence $\chi(AB^{p^{m-1}})\equiv\chi(A)\mod p^{m}$, establishing the claim. ∎

Let $L=\mathbb{Q}^{ur}_{p}$ be the maximal unramified extension of $\mathbb{Q}_{p}$. By Proposition 2.6, there is a $LG$-module $V$ affording $\chi$. Consider the radical filtration on $V$ defined by the action of $g$, and let $\operatorname{gr}(V)$ be the associated graded. Since $h$ commutes with $g$, the action of $h$ on $V$ induces an action of $h$ on $\operatorname{gr}(V)$.

Let $g=g_{s}g_{u}$ be the $p$-semisimple and $p$-unipotent decomposition of $g$. Since $g_{s}$ has order prime to $p$, all its eigenvalues lie in $L$, so $\operatorname{gr}(V)$ has a weight space decomposition with respect to the semisimple operator $g_{s}$. Let $W$ be a weight space for $g_{s}$. Since $g_{s}$ and $g_{u}$ are powers of $g$, $h$ commutes with $g_{s}$ and $g_{u}$. Hence $g_{u}$ and $h$ are commuting operators on $W$. By Lemma 3.3, $tr(g_{u}|W)\equiv tr(g_{u}h^{p^{m-1}}|W)\mod p^{m}$. Since $g_{s}$ is acting by a scalar on $W$, $tr(g|W)\equiv tr(gh^{p^{m-1}}|W)\mod p^{m}$. Summing over all weights of $g_{s}$ gives the desired result. ∎

Given $\mu$ and $\nu$ as in the statement, let

Observe that every $\rho\in RD(\lambda,\nu)$ has all rows of $(pk)^{p^{m-1}}$ placed in the row of length $n-pk$. This gives a bijection $R_{1}\cong RD(\lambda,\nu)$. Also, there are $(p^{m})!/(p^{m}-p)!$ ways of choosing $p$ out of $p^{m}$ rows with some order, and given such a choice, tiling the rows $k^{p}$ accordingly gives a unique element of $R_{3}$. It follows that

By Wilson’s theorem, $(p-1)!\equiv-1\mod p$, so $|R_{3}|\equiv-p^{m}\mod p^{m+1}$.

By Lemma 4.4, $|R_{1}|+|R_{2}|+|R_{3}|=|RD(\lambda,\mu)|$ and $|R_{1}|=|RD(\lambda,\nu)|$. We have shown $|R_{3}|\equiv-p^{m}\mod p^{m+1}$; thus, we must show $|R_{2}|\equiv 0\mod p^{m+1}$.

We show $p^{m+1}$ divides $|R_{2}|$ by considering orbits of a group and its $p$-Sylow subgroup acting on $R_{2}$. There is an action of the symmetric group $S_{p}$ on $RD(\lambda,\mu)$ by permuting the $p$ rows tiled into $k^{p}$. There is a commuting action of $S_{p^{m}}$ on $RD(\lambda,\mu)$ by permuting the placements of the rows of length $k$ of $\mu$ not in $\xi$. Given $\rho\in R_{2}$, let $Q$ be the $p$-Sylow subgroup of the stabilizer of $\rho$ in $S_{p}\times S_{p^{m}}$. We will analyze the size of $Q$.

Let $\pi_{1}:Q\to S_{p}$ and $\pi_{2}:Q\to S_{p^{m}}$ be the projections onto the two direct factors of $S_{p}\times S_{p^{m}}$. If $\sigma\in Q$, then since $\rho$ has between $1$ and $p-1$ rows of $k^{p^{m}}$ placed into $k^{p}$, $\pi_{1}(\sigma)$ does not act transitively on the $p$ rows in $k^{p}$. Since $Q$ is a $p$-group, $\pi_{1}(\sigma)$ is either the identity or the $p$-cycle; since $\pi_{1}(\sigma)$ does not act transitively, $\pi_{1}(\sigma)=1$. Thus $\pi_{1}$ is trivial and $\pi_{2}$ is injective.

If $\rho$ has exactly $\ell$ rows of $k^{p^{m}}$ placed into $k^{p}$, then $\pi_{2}(\sigma)$ must fix those $\ell$ rows. Hence $\pi_{2}(Q)\subseteq S_{p^{m}-\ell}$. Now $p^{m}$ divides $[S_{p^{m}}:S_{p^{m}-\ell}]$. It follows that $p^{m+1}$ divides $[S_{p}\times S_{p^{m}}:Q]$. As the index of $Q$ in the stabilizer of $\rho$ is prime to $p$, we conclude $p^{m+1}$ divides the cardinality of the orbit of $\rho$.

Since $p^{m+1}$ divides the size of every orbit of $S_{p}\times S_{p^{m}}$ on $R_{2}$, it follows that $p^{m+1}$ divides $|R_{2}|$. This proves the claim. ∎

i) $\implies$ ii): suppose that $\sigma\sim_{m}^{comb}\sigma^{\prime}$. We may assume that the cycle type of $\sigma^{\prime}$ is formed from that of $\sigma$ by replacing $p^{m}$ cycles of length $k$ with $p^{m-1}$ cycles of length $pk$. Write $\sigma=\sigma_{1}\sigma_{2}$ where $\sigma_{1}$ is a product of $p^{m}$ disjoint $k$-cycles and $\sigma_{2}$ is disjoint from $\sigma_{1}$. Then $\sigma_{1}=\tau^{p^{m}}$ for $\tau$ a $kp^{m}$-cycle also disjoint from $\sigma_{2}$. Since $\tau^{p^{m-1}}$ is a product of $p^{m-1}$ disjoint cycles of length $kp$, we see that the conjugacy class of $\sigma^{\prime}$ is represented by $\tau^{p^{m-1}}\sigma_{2}=\tau^{p^{m-1}-p^{m}}\sigma=(\tau^{p-1})^{p^{m-1}}\sigma$. Since $\tau$ commutes with $\sigma$, we obtain $\sigma\sim_{m}\sigma^{\prime}$.

ii) $\implies$ iii): this is Theorem 3.4.

iii) $\implies$ i): We proceed by double induction: on $m$, and backwards on the length of the smallest cycle where $\sigma$ and $\sigma^{\prime}$ differ in cycle type. The base case $m=1$ follows from Brauer’s theory: if $\sigma$ has cycle type $\mu$, then the semisimple part $\sigma_{s}$ has cycle type formed by replacing parts of size $p^{\ell}k,(k,p)=1$, in $\mu$ with $p^{\ell}$ parts of size $k$. Hence, if $\sigma_{s}$ and $\sigma^{\prime}_{s}$ are conjugate, then $\sigma\sim_{1}^{comb}\sigma^{\prime}$.

Now suppose that $m\geq 2$ and $\sigma,\sigma^{\prime}\in S_{n}$ are such that $\chi(\sigma)\equiv\chi(\sigma^{\prime})\mod p^{m}$ for all characters $\chi$ of $S_{n}$. Let $k$ be the smallest cycle length which appears in $\sigma$ and $\sigma^{\prime}$ with different multiplicity. Then $\sigma$ and $\sigma^{\prime}$ have cycle types $\mu=(\xi,k^{i},\eta)$ and $\nu=(\xi,k^{j},\tau)$ for partitions $\xi$, $\eta$, and $\tau$ such that all parts in $\eta$ and $\tau$ have size greater than $k$. By induction, we have $\sigma\sim_{m-1}^{comb}\sigma^{\prime}$, which implies $p^{m-1}\mid(i-j)$. This implies $n\geq pk$, so we may set $\lambda=(n-pk,k^{p})$. Any row decomposition of $\lambda$ by $\mu$ or $\nu$ places no parts of $\eta$ or $\tau$ into the parts of length $k$. Hence applying Lemma 4.5 $(i-j)/p^{m-1}$ times gives

Since $\chi(\sigma)\equiv\chi(\sigma^{\prime})\mod p^{m}$ for all characters $\chi$ of $S_{n}$, we conclude $p^{m}\mid(i-j)$. Assume without loss of generality that $j>i$. Then $\sigma^{\prime}\sim_{m}^{comb}\sigma^{\prime\prime}$ for $\sigma^{\prime\prime}$ of cycle type $(\xi,k^{i},(pk)^{(i-j)/p},\tau)$. By the case i) $\implies$ iii) above, we have $\chi(\sigma^{\prime})\equiv\chi(\sigma^{\prime\prime})\mod p^{m}$ for all characters $\chi$ of $S_{n}$. Now $\sigma$ and $\sigma^{\prime\prime}$ have the same number of cycles of length $k$, so by induction, $\sigma\sim_{m}^{comb}\sigma^{\prime\prime}$ and thus $\sigma\sim_{m}^{comb}\sigma^{\prime}$.

∎