Quatrième partie · À la poursuite des suturesChapitre 17

L’écart de Galois

The Galois gap

établiRead from the draft of 3 October 2026

Plate 17.1The character table of the group of order 168, with the sizes of its classes. A non-square power map exchanges the columns

7A

and

7B

; on the values it acts as Galois does,

\sqrt{-7}\mapsto-\sqrt{-7}

, so it exchanges the rows

\chi_3

and

\bar\chi_3

. The gold block holds the only irrational values.

17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8

Which twists of a group can no count of fixed points detect, and which of its characters can no finite set express?

Counting cannot tell the points of the Fano plane from its lines. Every element of the group of order 168 fixes as many points as lines, and yet no seam joins the two sets, since the stabilizer of a point fixes no line. Chapter 5 found the twist that exchanges them in arithmetic: it is the Galois conjugation of $\sqrt{-7}$ . This chapter shows that the two facts are one.

Counting is blind exactly to the twists that act on $G$ as Galois acts on its classes. In the other direction, finite $G$ -sets express exactly the characters that Galois fixes. One partition, the Galois orbits on the character table, measures both blindnesses, and what it leaves out, the number of conjugacy classes minus the number of rational classes, is the Galois gap. For the group of order 168 the gap is one-dimensional: it is the sign of $\sqrt{-7}$ .

The central result · What counting cannot hear

For an automorphism $\alpha$ of a finite group $G$ the following are equivalent.

(1) $\alpha$ is Galois-like.

(2) For every subgroup $H$ , the $G$ -sets $G/H$ and $G/\alpha(H)$ have the same permutation character: every $g\in G$ has as many fixed points on one as on the other, equivalently $\Q[G/H]\cong\Q[G/\alpha(H)]$ as $\Q[G]$ -modules.

(3) $\alpha$ fixes every rational-valued character of $G$ .

(4) $\alpha$ maps every irreducible character into its $\Gamma$ -orbit.

Proof

The permutation character of $G/H$ is $\pi_H(g)=|C_G(g)|\,|H\cap g^G|/|H|$ , so (2) says that $|H\cap C|=|\alpha(H)\cap C|$ for every class $C$ .

(1) $\Rightarrow$ (2). We have $|\alpha(H)\cap C|=|H\cap\alpha^{-1}(C)|$ , and $\alpha^{-1}(C)=C^{(k)}$ for some $k$ prime to the exponent $e$ . The map $x\mapsto x^k$ is a bijection from $H\cap C$ onto $H\cap C^{(k)}$ , with inverse $x\mapsto x^{k'}$ where $kk'\equiv1\pmod e$ . (2) $\Rightarrow$ (1). Take $H=\langle g\rangle$ . The group $\alpha(H)=\langle\alpha(g)\rangle$ meets the class of $\alpha(g)$ , so $\langle g\rangle$ meets it too. A conjugate of $\alpha(g)$ in $\langle g\rangle$ has the order of $g$ , so it is a generator $g^k$ .

(1) $\Leftrightarrow$ (3). Write $f^{(k)}(g)=f(g^k)$ for a class function $f$ . If $f=\sum c_\chi\chi$ is $\Gamma$ -invariant, comparing coefficients in $f^{(k)}=f$ gives $c_{\chi^{(k)}}=c_\chi$ , so the $\Gamma$ -invariant class functions are spanned by the $\Gamma$ -orbit sums of irreducible characters, which are rational-valued characters. They are the functions constant on rational classes, among them the indicator function of each rational class. Now $\alpha$ acts by $f\mapsto f\circ\alpha$ , and it fixes every function constant on rational classes exactly when it maps each rational class to itself.

(3) $\Leftrightarrow$ (4). $\alpha$ permutes the irreducible characters and commutes with $\Gamma$ . Since the irreducible characters are linearly independent, $\alpha$ fixes an orbit sum exactly when it maps the orbit to itself.

Status

Everything in the chapter is proved, and the table across the family is computed. The ingredients are classical. The implication from (1) to (2) is the usual way of producing Gassmann-equivalent subgroups from automorphisms; Sutherland uses the case where $\alpha$ fixes every class meeting $H$ , and the points and hyperplanes of a projective space are the classical instance, because the inverse transpose maps $g$ to a conjugate of $g^{-1}$ . The theorem on what finite sets cannot say is Artin’s induction theorem together with the Galois form of Brauer’s permutation lemma.

The equivalence of the four conditions, and the reading of the two theorems as one gap, were not found stated in the sources consulted. The content is elementary; its use here is to name exactly what counting is blind to.

Galois sur la table des caractèresGalois on the character table

Plate 17.1The character table of the group of order 168, with the sizes of its classes. A non-square power map exchanges the columns

7A

and

7B

; on the values it acts as Galois does,

\sqrt{-7}\mapsto-\sqrt{-7}

, so it exchanges the rows

\chi_3

and

\bar\chi_3

. The gold block holds the only irrational values.

Let $e$ be the exponent of the finite group $G$ , and let $k$ be prime to $e$ . The $k$ -th power map permutes the conjugacy classes, and the rule $\chi^{(k)}(g)=\chi(g^k)$ permutes the irreducible characters, because $\chi(g^k)=\sigma_k(\chi(g))$ for the element $\sigma_k\colon\zeta_e\mapsto\zeta_e^k$ of $\operatorname{Gal}(\Q(\zeta_e)/\Q)$ . So $\Gamma=(\Z/e\Z)^\times$ acts on the columns of the character table by power maps and on its rows by Galois conjugation, and the two actions are one.

For the group of order 168, $e=84$ . There are six classes, of sizes 1, 21, 56, 42, 24, 24, and six characters, of degrees 1, 3, 3, 6, 7, 8; $7A$ is the class of $z\mapsto z+1$ and $7B$ that of $z\mapsto z+3$ . The only irrational values are $\alpha=(-1+\sqrt{-7})/2$ and $\bar\alpha$ , taken by $\chi_3$ and $\bar\chi_3$ on $7A$ and $7B$ . A $k$ that is a square modulo 7 fixes every class and every character; a non-square exchanges $7A$ with $7B$ and, by the same stroke, $\chi_3$ with $\bar\chi_3$ . The rational classes are $\{1A\}$ , $\{2A\}$ , $\{3A\}$ , $\{4A\}$ and $\{7A,7B\}$ .

Definition

Let $e$ be the exponent of $G$ , and let $\Gamma=(\Z/e\Z)^\times$ act on the conjugacy classes by power maps, $C\mapsto C^{(k)}$ , the class of $g^k$ for $g\in C$ , and on the irreducible characters by $\chi^{(k)}(g)=\chi(g^k)=\sigma_k(\chi(g))$ , where $\sigma_k\colon\zeta_e\mapsto\zeta_e^k$ . A rational class is a $\Gamma$ -orbit of conjugacy classes, that is, the set of generators of the conjugates of one cyclic subgroup.

Les automorphismes de type galoisienGalois-like automorphisms

Plate 17.2The six classes, each with a representative, and the outer automorphism, conjugation by

z\mapsto-z

: it fixes four classes and exchanges

7A

and

7B

, so it sends every class into its rational class, as

g\mapsto g^{-1}

does.

A twist by an automorphism changes nothing a count can see when it moves each class only within its rational class. The definition names such automorphisms, and the twists they give are the inaudible ones.

At 168 the outer automorphism is conjugation by $z\mapsto-z$ , which lies in $\PGL(2,7)$ and not in $\PSL(2,7)$ , since $-1$ is not a square modulo 7. It fixes the classes $1A$ , $2A$ , $3A$ , $4A$ and exchanges $7A$ with $7B$ , the class of the inverses: $\alpha(z+1)=z-1=(z+1)^{-1}$ . So $\alpha(g)$ is conjugate to $g^{-1}$ for every $g$ , and the outer automorphism is Galois-like, realized on all classes at once by $k=-1$ , complex conjugation. In the life of the Fano plane the same class of automorphisms is the inverse transpose, which sends every element to a conjugate of its inverse.

Definition(Galois-like automorphisms)

An automorphism $\alpha$ of $G$ is Galois-like if it maps every conjugacy class into its rational class: for every $g\in G$ there is a $k$ prime to the order of $g$ with $\alpha(g)$ conjugate to $g^k$ . Inner automorphisms are Galois-like, and since automorphisms commute with power maps the Galois-like automorphisms form a normal subgroup of $\Aut(G)$ . Its image in $\operatorname{Out}(G)$ is the group of inaudible twists.

Ce que le dénombrement n’entend pasWhat counting cannot hear

Plate 17.3The fifteen objects on the line of their sizes, the

a

-classes above it and the

b

-classes below, which the outer automorphism exchanges. The three Gassmann pairs, bracketed, have the same counts and no seam.

The permutation character of $G/H$ is $\pi_H(g)=|C_G(g)|\,|H\cap g^G|/|H|$ , so two transitive $G$ -sets have the same counts exactly when their stabilizers meet every class in equally many elements, when they are Gassmann equivalent. A Galois-like twist preserves these numbers, since $x\mapsto x^k$ carries $H\cap C$ bijectively onto $H\cap C^{(k)}$ . Conversely a twist that is not Galois-like moves some element $g$ out of its rational class, and the cyclic subgroup $\langle g\rangle$ hears it. That is the central theorem, and its four conditions say the same thing about subgroups, about rational characters and about irreducible ones.

At 168 the outer automorphism moves exactly three pairs of objects: $G/V_4^a$ and $G/V_4^b$ , $G/A_4^a$ and $G/A_4^b$ , and $G/S_4^a$ and $G/S_4^b$ , the last being the points and the lines of the Fano plane. None of the groups $V_4$ , $A_4$ , $S_4$ contains an element of order 7, so each meets $7A$ and $7B$ in no element, and every count is the same on the two members of each pair. Their marks at $V_4^a$ still differ, so no seam joins them.

Proposition(The Gassmann pairs of

\PSL(2,7)

) proved

Two distinct objects of $G=\PSL(2,7)$ have the same permutation character exactly for the three pairs $(G/V_4^a,G/V_4^b)$ , $(G/A_4^a,G/A_4^b)$ and $(G/S_4^a,G/S_4^b)$ . Every bridge between the two members of such a pair, for one marking, is refuted.

Proof

The coincidences are read off the table of marks. Conceptually, the outer automorphism exchanges the two members of each pair and fixes every class of elements of $G$ except $7A$ and $7B$ ; the groups $V_4$ , $A_4$ , $S_4$ contain no element of order 7, so the two members meet each class equally often, which is the condition for equal permutation characters. This is an instance of the theorem on counting: the outer automorphism maps $7A$ to $7B=7A^{-1}$ . The two members of each pair are not isomorphic $G$ -sets, since their marks at $V_4^a$ differ, so no seam joins them and the bridge is refuted.

Ce que les ensembles finis ne peuvent direWhat finite sets cannot say

Plate 17.4The permutation characters of the fifteen objects on the six classes. Every row takes one value on

7A

and

7B

, the gold band, so the fifteen span five dimensions; the Gassmann pairs, bracketed, have equal rows. Under the rule is the one direction they miss,

\chi_3-\bar\chi_3

A finite $G$ -set is seen through its counts, and its counts are its permutation character, a class function with rational values. The theorem says which class functions finite sets reach in rational combination: those constant on rational classes, and nothing else.

At 168 the fifteen permutation characters, one for each object, take one value on $7A$ and $7B$ , and together they have rank 5: they span the class functions constant on the five rational classes. The sixth direction is $\chi_3-\bar\chi_3$ , zero on four classes and $\mp\sqrt{-7}$ on $7A$ and $7B$ . It is orthogonal to every permutation character, and no combination of $G$ -sets equals it.

Theorem(What finite sets cannot say) proved

Over $\Q$ , the permutation characters of the finite $G$ -sets span exactly the class functions that are constant on rational classes. The dimension of this span is the number of rational classes, which equals the number of $\Gamma$ -orbits on the irreducible characters. A complement in the space of class functions is spanned by the differences $\chi-\chi^{(k)}$ of Galois-conjugate irreducible characters. Its dimension, the number of conjugacy classes minus the number of rational classes, is the Galois gap of $G$ .

Proof

Permutation characters take rational values. By Artin’s induction theorem every rational-valued character is a rational combination of the permutation characters $1_C^G$ , $C$ cyclic. By the proof of the theorem on counting, the rational-valued characters span the functions constant on rational classes, and the orbit sums form a basis of that space. Averaging over $\Gamma$ projects onto it, and the kernel of the projection is spanned by the $f-f^{(k)}$ , hence by the $\chi-\chi^{(k)}$ .

Une partition, deux cécitésOne partition, two blindnesses

Plate 17.5Galois’s orbits on the six classes and on the six characters, five of each. The one pair on each side, gold, is the same conjugation: counting is blind to the twists that keep every class orbit, and finite sets reach only what is constant on the character orbits.

The partition is the same on both sides because the number of $\Gamma$ -orbits on classes equals the number on characters: both are the dimension of the $\Gamma$ -invariant class functions. At 168 there are five of each. The one pair of classes, $\{7A,7B\}$ , and the one pair of characters, $\{\chi_3,\bar\chi_3\}$ , are the same conjugation seen from two sides: the irrational characters differ only on the classes of order 7.

The two blindnesses are of different kinds. The first is a statement about twists: the outer automorphism of 168 keeps every orbit, so no count hears it. The second is a statement about sets: no combination of $G$ -sets reaches a function that separates $7A$ from $7B$ . Where the gap is zero, as in the Weyl family, finite sets express every character, and only a twist that fixes every class can be inaudible.

Corollary(One partition, two blindnesses) proved

The $\Gamma$ -orbits on the irreducible characters govern both theorems. Finite sets express exactly the combinations of characters that are constant along these orbits. Counting fails to hear exactly the twists that preserve each orbit. For every finite $G$ -set $X$ and every Galois-like $\alpha$ , the twisted set $X_\alpha$ has the permutation character of $X$ , because $\pi_X$ is constant on rational classes.

When all characters of $G$ are rational, the gap is zero: finite sets express every character, and a twist is inaudible only if it fixes every conjugacy class. This holds for the symmetric groups and for every finite Weyl group, so it holds for the Weyl family of Chapter 8.

L’écart à travers la familleThe gap across the family

Plate 17.6The twists across the family: whether each is inaudible, with the Galois element that realizes it; the pairs of subgroup classes it moves; and the gap. Only the two exotic twists of

A_6

are heard, and only they move a pair that is not Gassmann.

$A_5=\PSL(2,5)$ : the twist from $\PGL(2,5)$ is inaudible, realized by $\sqrt5\mapsto-\sqrt5$ , it moves no pair of subgroup classes, and the gap is 1. $\PSL(2,7)$ : inaudible by complex conjugation, moving $V_4$ , $A_4$ and $S_4$ , all Gassmann; gap 1. $\SL(2,7)$ : the twist from $\GL(2,7)$ is inaudible by complex conjugation, moving the lifts of those three, of orders 8, 24 and 48, all Gassmann; gap 3. $\PSL(2,11)$ : inaudible by complex conjugation, moving $S_3$ and $A_5$ , both Gassmann; gap 2. $\PSL(2,8)$ : the field twist, of order 3, is inaudible by a Galois element of order 3 and moves nothing; gap 4.

$A_6=\PSL(2,9)$ is the instructive case, with three twists. The field twist, which $S_6$ realizes, is inaudible, by $\sqrt5\mapsto-\sqrt5$ , and moves no pair; the gap is 1. The diagonal twist, from $\PGL(2,9)$ , and their product, from $M_{10}$ , are audible: each moves six pairs of subgroup classes, $C_3$ , $V_4$ , $S_3$ , $A_4$ , $S_4$ and $A_5$ , and only the $V_4$ pair is Gassmann. They exchange 3-cycles with products of two 3-cycles, and the cyclic groups of order 3 witness it.

Proposition(The gap across the family) computed

For the groups of the double lives and of the trinity, and for $\SL(2,7)$ , the twists generating $\operatorname{Out}(G)$ behave as in the table. In every case, Galois-like holds exactly when all moved pairs of subgroup classes are Gassmann equivalent, and exactly when the twist maps every irreducible character into its Galois orbit.

Proof

Computed by machine with permutations, all subgroup classes enumerated, and the characters found by Burnside’s algorithm and checked by orthogonality.

Le signe de √−7 ne se compte pasThe sign of √−7 cannot be counted

Plate 17.7The character table again, with the row

\chi_3-\bar\chi_3

beneath it: zero on four classes and

\mp\sqrt{-7}

7A

and

7B

. It spans the gap, and no combination of finite

G

-sets reaches it.

Every count made from the group, the number of fixed points of any element on any finite set built from it, is the same whichever square root of $-7$ is called $\sqrt{-7}$ , and no combination of such sets separates $\chi_3$ from $\bar\chi_3$ . Fixing the sign takes a datum that is not a count: a choice of $\sqrt{-7}$ , the rule that Chapter 5 found deciding which orbit of seven conics is the points of the Fano plane.

On the double cover the gap is three: $\SL(2,7)$ has eleven classes and only eight rational classes. The difference of the Weil quartet and its conjugate lies in the gap, so no combination of finite $\SL(2,7)$ -sets separates the two either.

Corollary(The sign of

\sqrt{-7}

cannot be counted) proved

For $G=\PSL(2,7)$ the gap is one-dimensional. It is spanned by $\chi_3-\bar\chi_3$ , whose values are $\pm\sqrt{-7}$ on $7A$ and $7B$ and 0 elsewhere. The outer automorphism acts on the irreducible characters as complex conjugation, so it is inaudible.

No count tells the points of the Fano plane from its lines, or $G/V_4^a$ from $G/V_4^b$ , or $G/A_4^a$ from $G/A_4^b$ . No combination of $G$ -sets equals $\chi_3-\bar\chi_3$ .

For the double cover $\SL(2,7)$ the gap is three-dimensional, spanned by $\chi_3-\bar\chi_3$ , $\chi_4-\bar\chi_4$ and the difference of the two faithful characters of degree 6. The characters $\chi_4$ and $\bar\chi_4$ are the Weil quartet and its conjugate. The two characters of degree 6 are exchanged by $\sqrt2\mapsto-\sqrt2$ . The outer automorphism again acts as complex conjugation, so it exchanges the quartets and fixes the last pair.

VérificationsThe checks

Plate 17.8

A_6

heard. On six letters, the cosets of one class of

A_5

, an element of order 3 is a 3-cycle fixing three letters; the field twist, from

S_6

, sends it to another 3-cycle, and the diagonal twist, from

\PGL(2,9)

, to two 3-cycles fixing none.

The theorem says which twists a count can hear, and the family checks it in both directions. $A_6$ makes the audible case visible. On the six cosets of one class of $A_5$ , an element $g$ of order 3 is a 3-cycle and fixes three letters. The field twist sends it to another 3-cycle, which fixes three letters too; the diagonal twist sends it to a product of two 3-cycles, which fixes none. Counting fixed letters hears the diagonal twist, and $\langle g\rangle$ is the cyclic witness the proof asks for.

Seven letters give an inaudible case beyond the family: the two classes of Fano planes in $A_7$ have the same counts and are not conjugate, and the twist between them acts on the classes as $g\mapsto g^{-1}$ does. The gap was computed for $\PSL(2,q)$ with $q=5$ , 7, 8, 9, 11, for $\SL(2,7)$ and for the two classes of $\PSL(3,2)$ in $A_7$ , with all subgroup classes enumerated by closure under joins, the outer automorphisms realized in $\mathrm{P\Gamma L}(2,q)$ , and the characters found by Burnside’s algorithm and checked by orthogonality.

Remark(Seven letters) computed

The alternating group $A_7$ has two classes of subgroups $\PSL(3,2)$ , the two $A_7$ -orbits of fifteen Fano planes on seven letters. They are exchanged by $S_7$ , and they are Gassmann equivalent. The twist exchanges $7A$ with $7B$ , the class of the inverses, and fixes every other class, so it is Galois-like.

The gap names exactly what counting is blind to. At 168 it is a single sign, and every count made from the group is the same for both values of it. Where the gap is zero, as for the symmetric groups and the Weyl family, finite sets express every character.

The gap is a statement about blindness, not about choice. Which identifications a theory makes naturally, and whether the twists such identifications can pick up are Galois symmetries of an arithmetic source, are the questions of the reciprocity law of Chapter 16.