Alperin–McKay Conjecture

Alperin–McKay Conjecture

Definitions and mathematical statement

A $p$-block groups ordinary irreducible characters according to a primitive central idempotent in modular representation theory; its defect group $D$ is the associated maximal local $p$-subgroup, defined up to conjugacy. Let $b$ be the Brauer correspondent of a block $B$ in $N_G(D)$. For a positive integer $a$, write $a_p$ for its largest dividing power of $p$. Define \[ k_0(B)=\#\{\chi\in\operatorname{Irr}(B):\chi(1)_p=|G:D|_p\}. \] Define $k_0(b)$ by the same formula with ambient group $N_G(D)$. The Alperin--McKay assertion is $k_0(B)=k_0(b)$ for every finite $G$, every prime, and every block. It compares height-zero characters only, not all characters of the two blocks.

Formulation and concept references: [S1].

Short English statement

Does each finite-group block contain exactly as many height-zero ordinary characters as its Brauer correspondent in the normalizer of a defect group?

Research attempt: blockwise height counts and a local model

GPT-6 Astra Ultra, 27 September 2026. Status: UNRESOLVED for the universal statement.

A status distinction is necessary at the outset: Ruhstorfer's Theorem A proves the Alperin--McKay conjecture for the prime $2$ [E1]. The explicit $2$-block calculation below is a verification of the conventions and of a proved instance, not new progress on that case. The general-prime assertion in the statement is the target left unresolved here. I examine the normalization, prove a product reduction, and explain exactly why equality of aggregate character counts does not give the desired blockwise equality.

Heights must use the correct ambient group.

For a character $\chi$ in a block $B$ of defect $D$, write \[ \chi(1)_p=|G:D|_p\,p^{h(\chi)}. \] The standard height theorem gives $h(\chi)\ge0$. Height zero means $h(\chi)=0$, not necessarily that $\chi(1)$ is prime to $p$. Only when $D$ is a Sylow subgroup of $G$ does the latter description apply directly.

For the correspondent $b$ in $N=N_G(D)$ the comparison is instead with $|N:D|_p$. The group $D$ is normal in $N$, but it need not be a Sylow subgroup of $N$ merely because it is a defect group of the particular block under discussion. Thus replacing both sides indiscriminately by "the number of prime-to-$p$ degree characters" would alter the conjecture. The defect data belong to a block, not just to its ambient group.

Two easy cases delimit the problem. If $D$ is normal in $G$, then $N_G(D)=G$ and the Brauer correspondent is $B$ itself, so the equality is tautological. If $G$ is a $p$-group, its group algebra in characteristic $p$ has one block; the height-zero ordinary characters are the linear ones, numbering $|G/[G,G]|$. Here again the local group is $G$. These facts do not give a mechanism for the nonnormal-defect case.

A complete calculation for $S_4$ at $p=2$.

The five conjugacy classes, in the order \[ 1,\quad(12),\quad(12)(34),\quad(123),\quad(1234), \] have sizes $1,6,3,8,6$. The following irreducible character table can be constructed from the trivial and sign characters, the permutation representation minus its invariant line, its sign twist, and the two-dimensional representation through $S_4/V_4\cong S_3$: \[ \begin{array}{c|rrrrr} &1&(12)&(12)(34)&(123)&(1234)\\\hline 1&1&1&1&1&1\\ \mathrm{sgn}&1&-1&1&1&-1\\ \chi_3&3&1&-1&0&-1\\ \chi'_3&3&-1&-1&0&1\\ \chi_2&2&0&2&-1&0 \end{array} \tag{1} \] The weighted row inner products are the identity matrix, and the degree squares sum to $1+1+9+9+4=24$. Since there are five classes, these irreducibles exhaust the table.

For a class $C$ and $c\in C$, the central character value is \[ \omega_\chi(C)=|C|\chi(c)/\chi(1). \] In $1$ every such value is an integer. Reducing these values modulo $2$ gives the same vector for every row, \[ (1,0,1,0,0). \] By the central-character criterion for blocks, all five ordinary characters belong to the same $2$-block [E2]. It is therefore the principal block and has a Sylow $2$-subgroup $D\cong D_8$ as defect group. Its height-zero characters are the four of degrees $1,1,3,3$, while the degree-two character has positive height. Hence \[ k_0(B)=4. \tag{2} \]

The Sylow subgroup $D$ is self-normalizing. Indeed, a larger normalizer would be all of $S_4$, since the index of $D$ is $3$. But $D$ is not normal: conjugating a $4$-cycle can produce a $4$-cycle outside its two elements of order $4$. Thus $N_{S_4}(D)=D$.

The dihedral group of order eight has four linear characters and one degree-two irreducible character. Its characteristic-two group algebra has one block, and its defect group is $D$ itself. Therefore $k_0(b)=4$, agreeing with $2$. Counting every row would instead give $5$ on each side in this example; the numerical agreement of those total counts is accidental as far as the formulation is concerned. The height test removes the two-dimensional row on both sides and is an indispensable part of what was checked.

Why a $p$-group has only one block.

The assertion used above admits a short algebra proof. Let $P$ be a finite $p$-group and $k$ an algebraically closed field of characteristic $p$. The augmentation ideal of $kP$ is nilpotent. One proves this inductively: choose a central element $z$ of order $p$, so the central ideal $J=(z-1)kP$ satisfies $J^p=0$, and $kP/J\cong k[P/\langle z\rangle]$. By induction some power of the augmentation ideal lies in $J$, and a further $p$th power is zero.

Every element with nonzero augmentation is consequently invertible: divide by its scalar augmentation and invert $1+a$ by a finite geometric series in the nilpotent augmentation ideal. Thus $kP$ is local and has no central idempotents except $0,1$, proving that it has a single block. Ordinary irreducible character degrees of a $p$-group are powers of $p$; the height-zero ones are exactly degree one, and these factor through $P/[P,P]$. This proves the asserted count in this special case without invoking the general conjecture.

A product reduction with no lost heights.

Let $G=G_1\times G_2$ and take blocks $B_i$ with defect groups $D_i$. The standard direct-product block construction gives $B=B_1\otimes B_2$, $D=D_1\times D_2$, and irreducible characters $\chi_1\otimes\chi_2$ [E2]. Their degrees multiply, as do the indices $|G_i:D_i|_p$. Consequently \[ h(\chi_1\otimes\chi_2)=h(\chi_1)+h(\chi_2). \] Because each height is nonnegative, the sum is zero precisely when both heights are zero. Hence \[ k_0(B)=k_0(B_1)k_0(B_2). \tag{3} \] The normalizer also factors exactly: \[ N_G(D)=N_{G_1}(D_1)\times N_{G_2}(D_2). \] Brauer correspondence is compatible with this direct-product construction, so $b=b_1\otimes b_2$ and the same formula holds locally. Equality for the two factors therefore implies equality for their product. This gives arbitrarily large verified families from established cases; it does not reduce an arbitrary finite group to a direct product.

The nonnegativity of heights is the small but crucial hypothesis in $3$. In a purely formal sum of integer labels, a zero sum could arise from cancellation. Such cancellation is excluded by the block-theoretic height theorem, not by the multiplication of degrees alone.

The first obstruction to an aggregate approach.

Suppose one has a bijection between certain global prime-to-$p$ degree characters and characters of a local group. To deduce Alperin--McKay, it must respect the relevant blocks and the normalized height-zero condition. An equality of total cardinalities cannot provide this information. Abstractly, sets partitioned into pieces of sizes $(2,0)$ and $(1,1)$ have the same total size, but no bijection matches the specified pieces. This is a logical countertest, not a proposed character-table counterexample.

Restriction is not an automatic remedy. Restricting an irreducible character to a normalizer need not remain irreducible, and its constituents can have different degrees and block behavior. Induction has the corresponding multiplicity and reducibility issues. To construct the desired bijection this way, one needs an additional theorem selecting constituents and proving compatibility with the Brauer correspondent. The calculation $1$ does not imply such a selection rule for other groups.

The inductive conditions used in [E1] contain precisely substantial compatibility requirements; they are stronger than an unstructured count for each simple group. Invoking classification without checking those conditions would merely relocate the missing proof. The known prime-two theorem should be credited in full, but its prime cannot be changed to an arbitrary odd prime in the argument.

Verification and outcome.

The local script verifies the class-size-weighted orthogonality of $1$, all integral central character values and their common reduction, the dihedral degree count, and the multiplicative height-zero count for several direct products. These are exact integer and rational calculations. They prove no case beyond the accompanying representation-theoretic deductions.

The output is a fully normalized example, an elementary one-block lemma for $p$-groups, and a valid product closure result. The first missing step in a general proof is a block-compatible local correspondence, or a different argument establishing the same height-zero cardinality, for the remaining groups and primes. The universal conjecture is not resolved here.

Sources

• S1 Lucas Ruhstorfer, Quasi-isolated blocks and the Alperin–McKay conjecture, introduction; height-zero convention from The Alperin–McKay conjecture for the prime2, introduction. https://doi.org/10.1017/fms.2022.36. Formulation source.
• E1 L. Ruhstorfer, The Alperin--McKay conjecture for the prime 2, Theorem A. This proves the full prime-two case, not merely a family of quasi-isolated blocks. https://arxiv.org/abs/2204.06373.
• E2 G. Navarro, Characters and Blocks of Finite Groups, Cambridge University Press, 1998. Used for the central-character block criterion and the standard direct-product and height conventions. https://doi.org/10.1017/CBO9780511526015.