Verification of the ordinary character table of the Baby Monster

TL;DR

This paper confirms the accuracy of the character table for the Baby Monster group, a large sporadic simple group, as presented in the Atlas of Finite Groups, ensuring its correctness for mathematical reference.

Contribution

It provides a rigorous proof verifying the correctness of the previously published character table of the Baby Monster group.

Findings

Confirmed the character table matches the Atlas data

Ensured the table's correctness for future research

Validated the structure of the Baby Monster group

Abstract

We prove the correctness of the character table of the sporadic simple Baby Monster group that is shown in the Atlas of Finite Groups.

Tables

Table 1. Table 1. Words to express t 1 , … , t 11 subscript 𝑡 1 … subscript 𝑡 11 t_{1},\ldots,t_{11} in terms of a , b 𝑎 𝑏 a,b

Table 2. Table 2. The names of our words

Equations

$$\overline{f}$$

$$\overline{g}$$

$$\overline{l}$$

$$\overline{l_3}$$

$$\overline{l_4}$$

$$a^{\prime }$$

$$b^{\prime }$$

$$\begin{array}[]{c|cccccccc|}&3A&3B&5A&5B&9A&9B&15A&15B\cr\hline\cr t_{2}(x)&1&0&0&1&0&1&0&1\cr\hline\cr\end{array}$$

$$\begin{array}[]{ccccccccccccc}\hline\cr 2A&2B&2C&2D&6A&6B&6C&6D&6E&6F\cr 1860&2048&2158&2168&3486&3510&3566&3534&3606&3604\cr\hline\cr 6G&6H&6I&6J&6K&10A&10B&10C&10D&10E\cr 3596&3610&3636&3638&3634&3860&3896&3918&3908&3920\cr\hline\cr 10F&14A&14B&14C&14D&14E&18A&18B&18C&18D\cr 3932&3996&4008&4048&4034&4052&4088&4090&4110&4124\cr\hline\cr 18E&18F&22A&22B&26A&26B&30A/B&30C&30D&30E\cr 4128&4122&4140&4158&4198&4176&4190&4212&4206&4214\cr\hline\cr 30F&30GH&34A&34BC&42A/B&42C\cr 4224&4216&4238&4220&4242&4258\cr\hline\cr\end{array}$$

$$\begin{array}[]{cccccc}\hline\cr 36A&36B&36C&60A&60B&60C\cr 4226&4238&4248&4280&4286&4296\cr\hline\cr\end{array}$$

$$\begin{array}[]{cccccccc}\hline\cr 28A/C&28B/D&28E&32AB/CD&40A/B/C&40D&40E&48A/B\cr 4188&4200&4210&4222&4242&4250&4258&4266\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|cccccc}&28A&28C&28B&28D&32AB&32CD\cr\hline\cr t_{3}(x^{2})&&&&&2&0\cr t_{3}(x^{7})&0&1&1&0\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|cccccc}&40A&40B&40C&48A&48B\cr\hline\cr t_{3}(x)&0&1&2&0&1\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|cccccccccc|}&4A&4B&4C&4D&4E&4F&4G&4H&4I&4J\cr\hline\cr r(1+x)&3114&3114&3192&3192&3256&3202&3204&3266&3264&3266\cr t_{3}(x)&1&0&1&0&&&&0&&1\cr r(1+x)^{3}&&&&&&&&1082&&1084\cr\hline\cr\end{array}$$

$$\begin{array}[]{ccccccccccc}\hline\cr 8A&8B/C/E&8D&8F&8H&8G&8I&8L&8J&8K/M&8N\cr 3774&3738&3778&3780&3780&3810&3786&3786&3812&3818&3818\cr&&&&&&3202&3204\cr&&&2619&2620&&&&&2714&2717\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccccc|}&8B&8C&8E&8K&8M\cr\hline\cr t_{3}(x)&1&0&2&1&2\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccccccc|}&12A/C/D&12B&12E&12F&12G/H&12I&12J\cr\hline\cr r(1+x)&3936&3942&3958&3996&3962&3964&3986\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccccccc|}&12K/M&12L&12N&12O&12P&12Q/R/T&12S\cr\hline\cr r(1+x)&3978&3966&4000&3982&3988&4002&4004\cr\hline\cr\end{array}$$

$$\begin{array}[]{r|ccc|cc|cc|ccc|}&12A&12C&12D&12G&12H&12K&12M&12Q&12R&12T\cr\hline\cr t_{3}(x)&0&1&0&1&0&0&1&0&1&2\cr t_{5}(x)&3&4&1&&&&&&&\cr\hline\cr\end{array}$$

$$\begin{array}[]{cccccccc}\hline\cr 24A/B/C/D&24E/G&24F&24H&24I/M&24J&24K&24L/N\cr 4152&4164&4170&4182&4176&4178&4174&4186\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|cccccc|}&24A&24B&24C&24D&24L&24N\cr\hline\cr t_{3}(x)&1&0&2&2&1&2\cr t_{5}(x)&3&0&2&1&&\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccc|}&16A/B&16C/D/E/F&16G/H\cr\hline\cr r(1+x)&4072&4074&4094\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccccccc|}&20A/B/C/D&20E&20F&20G&20H&20I&20J\cr\hline\cr r(1+x)&4114&4128&4132&4148&4144&4138&4150\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|cccc|}&20A&20B&20C&20D\cr\hline\cr r(1+x^{2})&3896&3908&3896&3896\cr t_{3}(x)&2&1&0&1\cr\hline\cr\end{array}$$

$$\begin{array}[]{c|ccccccc|}&16A&16B&16C&16E&16D/F&16G&16H\cr\hline\cr t_{3}(x)&0&1&2&2&0&&\cr t_{3}(x^{2})&&&&&&1&2\cr r(1+x^{2})&&&3780&3778&&&\cr\hline\cr\end{array}$$

$$\begin{array}[]{cccccccccc}2&3&4&5&6&7&175&176&177&178\cr 2a&2b&2c&2d&2e&2f&2g&2h&2i&2j\cr-1A&+2A&-2A&+2B&-2B&2C&+2D&-2D&+2E&-2E\end{array}$$

Full text

MSC: 20C15,20C40,20D08

In memory of our friend and colleague Kay Magaard, who sadly passed away during the preparation of this paper.

1. Introduction

Jean-Pierre Serre has raised the question of verification of the ordinary character tables that are shown in the $\mathbb{ATLAS}$ of Finite Groups [6]. This question was partially answered in the paper [5], the remaining open cases being the largest two sporadic simple groups, the Baby Monster group ${\mathbb{B}}$ and the Monster Group ${\mathbb{M}}$, and the double cover $2.{\mathbb{B}}$ of ${\mathbb{B}}$.

The current paper describes a verification of the character table of ${\mathbb{B}}$. The computations shown in [3] then imply that also the $\mathbb{ATLAS}$ character table of $2.{\mathbb{B}}$ is correct. As in [5], one of our aims is to provide the necessary data in a way that makes it easy to reproduce our computations.

The $\mathbb{ATLAS}$ character table of the Baby Monster derives from the original calculation of the conjugacy classes and rational character table by David Hunt, described very briefly in [9]. The irrationalities were calculated by the CAS team in Aachen [11].

2. Strategy

We begin with a preliminary section, Section 3, whose aim is to prove that certain specified matrices do indeed generate copies of the Baby Monster. These matrices can then be used in the main computation. The $Y_{555}$ presentation of the BiMonster implies a $Y_{433}$ presentation for ${\mathbb{B}}$ (see [10]), given that the Schur multiplier $H^{2}({\mathbb{M}},{\mathbb{C}}^{*})$ of the Monster has odd order. The Schur multiplier of the Monster was calculated by Griess [8]. We use the $Y_{433}$ presentation to prove that three pairs of matrices, of dimension $4370$ over the field with two elements and of dimension $4371$ over the fields with three and five elements, respectively, generate the group ${\mathbb{B}}$, and, moreover, that mapping one pair of these generators to any other such pair defines a group isomorphism.

In Section 4 we compute a first approximation to the list of conjugacy class names, by establishing invariants, in terms of the above three matrix representations of ${\mathbb{B}}$, that in fact distinguish almost all conjugacy classes of cyclic subgroups of ${\mathbb{B}}$. These invariants are then used to determine the power maps between the specified unions of conjugacy classes.

In order to compute the conjugacy classes of ${\mathbb{B}}$ and the corresponding centralizer orders, we apply the following general statement. For a group $G$ and an element $g\in G$, if $x,y\in G$ power to $g$ then $x\sim_{G}y$ if and only if $x\sim_{N}y$ where $N:=N_{G}(\langle g\rangle)$; moreover $C_{G}(x)=C_{N}(x)$. This implies that it suffices to find the normalizers (or overgroups thereof) of prime order subgroups, and their character tables. Section 5 deals with the first problem, Section 6 with the second.

At this point, we know the conjugacy classes of ${\mathbb{B}}$, and their lengths. Some further calculations then match these up with the names listed in Section 4, by using the given invariants and some small extra arguments. Finally, the irreducible characters of ${\mathbb{B}}$ are computed in Section 7, using character theoretic methods such as induction from several subgroups of ${\mathbb{B}}$.

3. Verifying a presentation for the Baby Monster

In this section we give words in the ‘standard generators’ for the Baby Monster, that represent the $11$ transpositions in the $Y_{433}$ presentation. This provides a relatively straightforward test to prove that a given black-box group is in fact isomorphic to the Baby Monster.

3.1. The presentation

A presentation for the Baby Monster sporadic simple group ${\mathbb{B}}$ was conjectured in the $\mathbb{ATLAS}$ [6], and proved by Ivanov [10], subject to the Monster not having a proper double cover. This hypothesis has been proved by Griess [8].

The presentation is on $11$ generators $t_{i}$ ($1\leq i\leq 11$), satisfying the Coxeter relations $t_{i}^{2}=1$ for all $i$, $(t_{i}t_{j})^{3}=1$ for $(i,j)=(1,2)$, $(2,3)$, $(3,4)$, $(4,5)$, $(5,6)$, $(6,7)$, $(7,8)$, $(5,9)$, $(9,10)$, $(10,11)$, and $(t_{i}t_{j})^{2}=1$ for $i<j$ otherwise. Adjoining one extra relation, $(t_{5}t_{4}t_{3}t_{5}t_{6}t_{7}t_{5}t_{9}t_{10})^{10}=1$, nicknamed the ‘spider relation’, gives a presentation for $2\times 2{}^{\textstyle.}{\mathbb{B}}$. To obtain a presentation for ${\mathbb{B}}$ itself, we need two extra relations, $(t_{5}t_{4}t_{3}t_{6}t_{7}t_{8}t_{9})^{9}=1$ and $(t_{5}t_{4}t_{3}t_{6}t_{9}t_{10}t_{11})^{9}=1$. Since the Coxeter diagram has three ‘arms’, of lengths $4,3,3$, this presentation is known as the $Y_{433}$ presentation.

Matrices generating a copy of the Baby Monster were first produced in the early 1990s [18]. These act on a vector space of dimension $4370$ over the field of order $2$. In order to prove, without relying on the character table, that these matrices do indeed generate the Baby Monster, a method was given for producing elements of this group that satisfy the $Y_{433}$ presentation for the Baby Monster. However, actual words for these elements were not given in [18].

In this section we rectify this deficiency in [18], and hence enable the reader to check relatively easily that the matrices given in [21], that are claimed to generate the Baby Monster in various different representations, do in fact generate the Baby Monster. In addition to the representation over the field of order $2$, already mentioned, we checked the representations over the fields of order $3$ and $5$ constructed in [14]. All three of these representations will be required later on, for determining certain class fusions and power maps.

We begin with the ‘standard generators’ in the sense of [19, 21], that is an element $a\in 2C$ and an element $b\in 3A$ such that $ab$ has order $55$ and $(ab)^{4}bab(ab^{2})^{2}$ has order $23$. The cited references explain how to find such generators in a group which is in fact isomorphic to the Baby Monster. All calculations described in this section were performed using the C Meataxe written by Michael Ringe [15], based on the original Meataxe of Richard Parker [13].

3.2. Finding the generators for the presentation

The calculations in this section were performed using the standard generators for (a group that is claimed to be) the Baby Monster in its $4370$-dimensional representation over the field of order $2$, taken from [21]. Following the 10-step method described in [18], we proceed as follows. Steps 1–4 are devoted to finding generators $t_{1},\ldots,t_{8},t_{11}$ for a particular subgroup $2\times S_{9}$. In Steps 5–8 we centralize successively the elements $t_{1}$, $t_{3}t_{4}$, $t_{6}t_{7}$ and $t_{8}$ to produce a small number of candidates for $t_{9}$ and $t_{10}$. These candidates are tested in Step 9, at which point all the required generators have been found. Step 10 tests the relations.

Step 1

Take an arbitrary $2A$-element, and call it $t_{11}$. Find $C(t_{11})\cong 2{}^{\textstyle.}{}^{2}E_{6}(2){:}2$.

The element $d=(ab)^{15}b$ has order $38$, and powers to the involution $t_{11}=d^{19}$. The centralizer of $t_{11}$ is generated by $d$ and $c=(at_{11})^{3}$.

Step 2

Find a subgroup $F\cong Fi_{22}{:}2$ inside $C(t_{11})$.

We restrict the representation of ${\mathbb{B}}$ to $H:=\langle c,d\rangle\cong 2{}^{\textstyle.}{}^{2}E_{6}(2){:}2$, find the composition factors using the chop program of the Meataxe, and extract a $78$-dimensional irreducible representation of the quotient $\overline{H}:={}^{2}E_{6}(2){:}2$, in which the computational searches for steps 2–4 are performed. (This use of a small representation reduces the computation time by a factor of around $10^{5}$.) The invariant $q(x)=\mathrm{rank}(1+x)$ is useful for identifying conjugacy classes. In particular, $\overline{cd^{5}}$ is an element of order $26$ in the outer half of $\overline{H}$, so powers to an involution $x\in 2D$, in the $\mathbb{ATLAS}$ notation for conjugacy classes in ${}^{2}E_{6}(2){:}2$. We calculate $q(x)=26$. As $\overline{c}$ is an involution in the outer half and $q(\overline{c})=36$, we deduce that $\overline{c}\in 2E$.

Similarly, $\overline{cd}$ has order $36$ and therefore powers to a $3C$-element $x$, with $q(x)=54$, while $\overline{cd^{2}}$ has order $48$ and therefore powers to a $3A$-element $y$, with $q(y)=42$. We then find that $\overline{cd^{3}}$ has order $30$ and powers to an element $z$ of order $3$ with $q(z)=48$. Hence $z$ is in class $3B$. Looking at a few groups generated by conjugates of $\overline{c}$ and $\overline{(cd^{3})^{10}}$ we quickly find that if $\overline{e}=\overline{((cd^{3})^{10})^{d}}$ then $F:=\langle\overline{c},\overline{e}\rangle\cong\mathrm{Fi}_{22}{:}2$.

Step 3

Find a subgroup $S\cong S_{10}$ inside $F$.

The elements $e$ and $cec$ then generate the subgroup $\mathrm{Fi}_{22}$ of index $2$ in $F$, in which we find $\overline{((ec)^{6}c(ec)^{3})^{2}(ec)^{2}cec}$ is an element of order $10$ powering to an element of $\mathrm{Fi}_{22}$-class $2A$. The element $\overline{(ec)^{8}c(ec)^{3}}$ has order $9$ and is most likely to be in class $9C$ in $\mathrm{Fi}_{22}$. Since there is no simple test for this, we proceed and hope for the best. Looking at conjugates of these elements we soon find a pair $f,g$ generating $S:=\langle f,g\rangle\cong S_{10}$, as follows.

[TABLE]

Step 4

In $\langle\overline{t_{11}}\rangle\times S$ find transpositions $\overline{t_{1}},\ldots,\overline{t_{8}}$ generating $T\cong S_{9}$ with the required Coxeter relations.

These transpositions can be taken as $\overline{t_{1}}=\overline{f}$ and $\overline{t_{n+2}}=\overline{f^{gfg^{n}}}$ for $0\leq n\leq 6$.

Step 5

Find $C=C(t_{1})\cong 2{}^{\textstyle.}{}^{2}E_{6}(2){:}2$.

This step has to be carried out in the $4370$ dimensional representation. Standard dihedral group methods give the element $p=((ab)^{5}t_{11}(ab)^{-5}t_{1}(ab)^{5})^{-1}$ that conjugates $t_{11}$ to $t_{1}$, so that $h=c^{p}$ and $i=d^{p}$ generate the group $C:=\langle h,i\rangle=C(t_{1})$.

Step 6

Find $D=C_{C^{\prime}}(t_{3}t_{4})\cong 6\times U_{6}(2)$.

Again we restrict the representation of ${\mathbb{B}}$ to $\langle h,i\rangle$, and extract a copy of the $78$-dimensional representation, in which to carry out steps 6–8. Following the instructions in [18] we found two elements of the centralizer of $\overline{t_{3}t_{4}}$ to be $\overline{j}=\overline{[(t_{5})^{i^{2}},t_{3}t_{4}]}$ and $\overline{k}=\overline{[(t_{5})^{i^{5}},t_{3}t_{4}]}$. Together with $\overline{t_{6}t_{7}}$ and $\overline{t_{8}}$, these are enough to generate $\overline{D}\cong 3\times U_{6}(2)$.

Step 7

Find $E=C_{D^{\prime}}(t_{6}t_{7})\cong 3\times U_{4}(2)$.

Similarly we found the following elements centralizing $\overline{t_{6}t_{7}}$:

[TABLE]

These are sufficient to generate $\overline{E}\cong 3\times U_{4}(2)$.

Step 8

Find the twelve [sic] transpositions in $E$ which commute with $t_{8}$.

There is a slight error in [18] at this point. There are in fact $13$ transpositions in $U_{4}(2)$ that centralize $t_{8}$, rather than $12$ as stated there. They are the $13$ transpositions in a copy of $2{}^{\textstyle.}(A_{4}\times A_{4}).2$. One is the central involution and the other $12$ are in the outer half. Together they generate a subgroup $2^{1+4}{:}S_{3}$ of index $3$. Presumably the central involution was omitted from the original calculation. However, it commutes with neither $t_{5}$ nor $t_{11}$, so is not a candidate for $t_{9}$ or $t_{10}$.

First we looked for conjugates of $\overline{l_{4}}$ that commute with $\overline{t_{8}}$, using the elements $\overline{l_{5}}=\overline{(ll_{3}l_{4})^{3}l_{3}l_{4}}$ and $\overline{l_{6}}=\overline{(ll_{3}l_{4})^{2}l_{3}l_{4}ll_{3}l_{4}}$, both of order $9$, for the conjugation. Then the four conjugates $\overline{m_{1}}=\overline{l_{4}}$, $\overline{m_{2}}=\overline{l_{4}^{l_{5}^{4}}}$, $\overline{m_{3}}=\overline{l_{4}^{l_{5}^{5}}}$ and $\overline{m_{4}}=\overline{l_{4}^{l_{5}^{2}l_{6}^{2}}}$ are sufficient to generate $2^{1+4}{:}S_{3}$. The central involution is $\overline{(m_{1}m_{2}m_{3}m_{4})^{3}}$. Then $\overline{m_{1}}$ and its conjugates by $\overline{m_{2}m_{3}}$, $\overline{m_{2}m_{4}}$, and $\overline{m_{3}m_{4}}$ are four transpositions mapping to the same involution in the quotient $S_{3}$, so give the full set of $12$ after conjugating by $\overline{m_{1}m_{2}}$ and $\overline{m_{2}m_{1}}$.

Step 9

Check these twelve [sic] transpositions for candidates for $t_{9}$ and $t_{10}$. There is only one possibility up to an obvious inner automorphism.

This step was again carried out in the $4370$ dimensional representation. The calculations that do not involve $t_{2}$ could have been done in $78$ dimensions, but the time saved would be of the order of one minute, so insignificant. Of the $13$ transpositions, the only one which commutes with $t_{2}$ and $t_{11}$ but not $t_{5}$ is $m_{1}^{m_{2}m_{3}m_{2}m_{1}}$. Hence this is the only possibility for $t_{9}$. There are two that commute with $t_{5}$ but not $t_{11}$, namely $m_{1}^{m_{2}m_{3}}$ and $m_{1}^{m_{2}m_{4}m_{1}m_{2}}$. But $t_{11}$ conjugates one to the other, so without loss of generality we may take $t_{10}=m_{1}^{m_{2}m_{3}}$.

3.3. Verifying the presentation

We now have a straight line program for producing the elements $t_{1},\ldots,t_{11}$ from the elements $a,b$. This program is given in Table 1 for convenience. It must be applied in every claimed representation of the Baby Monster, and then the relations of the presentation must be checked.

Step 10

Prove that $t_{1},\ldots,t_{11}$ satisfy all the required relations.

We check the $66$ Coxeter relations by finding the order of the elements $t_{i}t_{j}$ for all $j\geq i$. (The relations $t_{i}^{2}=1$ are implicit in the calculation, but were explicitly checked again.) Similarly the spider relation is checked by confirming that the element $t_{5}t_{4}t_{3}t_{5}t_{6}t_{7}t_{5}t_{9}t_{10}$ has order $10$. Finally, we check that $t_{5}t_{4}t_{3}t_{6}t_{7}t_{8}t_{9}$ and $t_{5}t_{4}t_{3}t_{6}t_{9}t_{10}t_{11}$ have order $9$. (In the particular representations we checked, this last check can be omitted, since it is straightforward to show in each case that the centre of the group is trivial. Indeed, Schur’s Lemma implies the centre consists of scalars, while the generators have determinant $1$ and the only scalar of determinant $1$ is $1$.)

We verified the relations in the three representations from [21], that is in dimension $4370$ over the field of order $2$, and in dimension $4371$ over the fields of orders $3$ and $5$. The total computation time was under $12$ hours.

3.4. Reversing the process

To complete the proof that the matrices given in [21] generate the Baby Monster, we have to reverse the process, and find the Atlas standard generators in terms of the $Y_{433}$ generators. First we make two elements $r=t_{1}t_{2}t_{3}t_{4}t_{5}t_{6}t_{7}t_{8}$ of order $9$ and $s=t_{5}t_{9}t_{10}t_{11}$ of order $5$. These elements in fact generate the whole group, but it is not necessary to prove this at this stage. Now $t_{1}t_{2}$ is an element in class $3A$. A random search produces the element $(r^{7}s)^{15}$ in class $2C$. (Again, it is not necessary to prove that this involution is in class $2C$. However, we used the conjugacy class invariants in [20] to guide us, and found that this element $x$ has $q(x)=2158$, which identifies the conjugacy class as $2C$.) Another random search gives a candidate pair of standard generators

[TABLE]

Finally, we use the chop program of the Meataxe to conjugate the matrices to a standard basis with respect to, first, the generators $a,b$, and then, the generators $a^{\prime},b^{\prime}$. This calculation must, of course, be carried out in each representation that we wish to check. We found that in all three of the representations in [21], the resulting pairs of matrices are identical, proving that all claimed generating sets do indeed generate the same group. (This does not mean that $a^{\prime},b^{\prime}$ are the same elements as $a,b$, merely that the pair $(a^{\prime},b^{\prime})$ is equivalent to $(a,b)$ under an automorphism of the group, and therefore under conjugation.) The total computation time was under $3$ hours.

4. Conjugacy class invariants and power maps

in the Baby Monster

4.1. Introduction

In this section, we produce a list of easily computed conjugacy class invariants for a specified list of elements of the Baby Monster, which are in fact good enough to distinguish all conjugacy classes of cyclic subgroups except $16D$ and $16F$. As a result, we have a splitting of the elements into small unions of conjugacy classes, and power maps between these unions of classes. The final splitting into conjugacy classes, and refinements of power maps, is done later.

In [20] (and also in [21]) there is a list of words in the Atlas standard generators of the Baby Monster, suitable powers of which are in fact representatives for the $184$ conjugacy classes. However, the proof given there depends on the accuracy of the $\mathbb{ATLAS}$ character table of the Baby Monster, and in particular on the accuracy of the power map information. It is therefore necessary to provide a new proof, which does not depend on the character table. We can of course use the words, as long as we do not quote from [20] any of the properties of the corresponding elements of the Baby Monster. We assume that the three representations of the Baby Monster given in [21] do indeed represent the Baby Monster. This was proved in Section 3.

4.2. The words and their names

In [20] there is a list of $76$ words for elements of specified orders, that in fact lie in the $76$ classes of maximal cyclic subgroups. There are in fact $175$ classes of cyclic subgroups altogether, including the trivial group. We can therefore take suitable powers of the $76$ words as a further set of $99$ words defining elements of the group.

First we label the $76$ words with the names given in [20]. These names will later, of course, be identified with unions of conjugacy classes, but at this stage they are simply names. We calculate the orders of the elements, and hence verify that the numerical part of the name is indeed the order of the element. We define our other $99$ words and their labels as the obvious powers from the first line to the second line of each row of Table 2.

At this stage, we have a list of $175$ words which give elements of the specified orders in the Baby Monster. Our job now is to find invariants that distinguish the alphabetical part of the name.

4.3. Invariants

We compute only the invariants that [20] tells us are useful. Besides the order, the invariants we use for an element $x$ are of the following types:

• •

mod $2$ type: the trace $t_{2}(y)$ and the rank $r(y)$ of selected polynomials $y=p(x)$ in $x$, in the mod $2$ representation;

• •

mod $3$ type: the trace $t_{3}(x^{k})$ of selected powers of $x$, in the mod $3$ representation;

• •

mod $5$ type: the trace $t_{5}(x)$ of $x$, in the mod $5$ representation.

The last two are expensive, and are only used when we know they will in fact be useful.

4.3.1. Odd-order elements

We find $24$ cyclic subgroups of odd-order elements. For orders $1$, $7$, $11$, $13$, $17$, $19$, $21$, $23$, $25$, $27$, $31$, $33$, $35$, $39$, $47$ and $55$, the only invariant we shall need is the order. For the other orders, $3$, $5$, $9$ and $15$, the trace in the $4370$ dimensional representation mod $2$ distinguishes two names in each case:

[TABLE]

4.3.2. Elements of twice odd order

For elements of order $38$, $46$, $66$ or $70$, no further invariant is required. In the remaining cases we compute the rank (or nullity) of $1+x$ on the $4370$ dimensional representation over the field of order $2$. This turns out to be a sufficient invariant to distinguish all cases except the elements of order $30$ and $42$. The rank of $1+x$ is tabulated below: note that in the case $26A$ the rank is given incorrectly in [20] as $4196$ instead of $4198$.

[TABLE]

The cases $30A$ and $30B$ can be distinguished by the rank of $1+x^{5}$, which is $3510$ and $3486$ respectively. The cases $42A$ and $42B$ can be distinguished by the rank of $1+x^{3}$, which is $3996$ and $4008$ respectively.

4.3.3. Elements of order at least $28$

For elements of order $44$, $52$, and $56$, no further invariant is required. For elements of order $36$ and $60$, the rank of $1+x$ is sufficient:

[TABLE]

For the remaining element orders, $28$, $32$, $40$, and $48$, we have the following values of the rank of $1+x$:

[TABLE]

In particular, this invariant is of no help for elements of order $32$ or $48$. All necessary cases can be separated by the trace mod $3$ of $x$ or $x^{2}$ or $x^{7}$:

[TABLE]

[TABLE]

4.3.4. Elements of order $4$ and $8$

The rank of $1+x$ distinguishes $7$ cases of elements of order $4$. Three of these split into two, according to the trace on the $4371$-dimensional representation mod $3$. An alternative invariant to distinguish $4H$ from $4J$ is the rank of $(1+x)^{3}$ in the mod $2$ representation.

[TABLE]

Similarly, for elements of order $8$, the rank of $1+x$ distinguishes $8$ cases, one of which is split by the rank of $(1+x)^{2}$, while the rank of $(1+x)^{3}$ splits two more:

[TABLE]

The trace modulo $3$ distinguishes the remaining cases

[TABLE]

4.3.5. Elements of order $12$ and $24$

The rank of $1+x$ distinguishes $14$ cases of elements of order $12$:

[TABLE]

[TABLE]

All except $12A/D$ can be split using the trace mod $3$. This last case would seem to require the trace mod $5$.

[TABLE]

Similarly for elements of order $24$, the rank of $1+x$ distinguishes $8$ cases

[TABLE]

Of these, we can distinguish $24I/M$ with the rank of $1+x^{2}$, which is $3986$ and $3982$ respectively, and $24E/G$ with the rank of $1+x^{3}$, which is $3774$ and $3778$ respectively. All the rest are distinguished by the trace mod $3$, apart from the case $24C/D$, which seems to require the trace mod $5$.

[TABLE]

4.3.6. Elements of orders $16,20$

In these cases, the rank of $1+x$ distinguishes the following:

[TABLE]

[TABLE]

For the elements of order $20$, the rank of $1+x^{2}$ distinguishes $20B$ from $20A/C/D$, which are distinguished from each other by the trace mod $3$:

[TABLE]

For the elements of order $16$, the trace mod $3$ distinguishes $16A/B$, and separates $16C/E$ from $16D/F$. Then $16C/E$ can be separated with the rank of $1+x^{2}$, and $16G/H$ with the trace of $x^{2}$ mod $3$.

[TABLE]

There would appear to be no easily computed invariant which distinguishes $16D$ from $16F$.

4.4. Checking the approximate power maps

We power up each of the given words, to every relevant power (that is, every power dividing the element order), and compute the necessary invariants of the resulting elements. We therefore know the power maps approximately. In every case the power maps agree with the character table in the $\mathbb{ATLAS}$ [6]. Indeed, some of the power maps form part of the definition of our set of class representatives, so the calculations in these cases can in fact be omitted. This includes the class $16D$, which is defined to be the square of the classes $32CD$. Hence it is not necessary to find an invariant to distinguish $16D$ from $16F$, in order to verify the power maps. All that remains in order to verify that the power maps are actually correct, is, firstly, to prove that the class list is correct, and secondly, to deal with any issues concerning algebraically conjugate classes. Details of the computations are given in [4].

5.

Centralizers of prime order elements in the Baby Monster

In this section we determine the classes of prime order elements, and the orders of their centralizers, in the Baby Monster. Much of this information comes from Stroth’s 1976 paper [16]. In cases where [16] does not give full information, our strategy is first to use a certified copy of the Baby Monster from [21] to give lower bounds on both the number of conjugacy classes and the orders of the respective centralizers, and then to use local arguments, together with information about the permutation representation on the $\{3,4\}$-transpositions, to show these are also upper bounds. For technical reasons, we deal with the primes in the order $2$, $3$, $7$, $17$, $11$, $13$, $19$, $23$, $5$, $47$, $31$.

5.1. Fusion of involutions

From [16] we see there are exactly four classes of involutions in ${\mathbb{B}}$, with representatives labelled $d$, $x_{36}(1)$, $d\beta$ and $dx_{33}(1)x_{32}(1)$ respectively. In the $\mathbb{ATLAS}$ [6], these are labelled $2A,2B,2C,2D$ respectively. The centralizers of $2A$ and $2B$ and $2C$ are given in detail in [16]. The centralizer order of $2D$ is given together with a rough description of the structure.

For the purposes of computation, it is necessary to match these classes to the names given in Section 4. Note that any element of order $38$ powers into class $2A$, and any element of order $22$ powers into class $2B$. An element of order $34$ powers into either $2A$ or $2C$. We can now compute the rank of $1+x$ for involutions $x$ in the certified copy of the Baby Monster in dimension $4370$ mod $2$, obtaining the values $1860$ for class $2A$ and $2048$ for class $2B$. For suitable $x$ obtained as the $17$th power of an element of order $34$, we obtain the value $2158$, and for another involution we obtain $2168$, so these are in class $2C$ and $2D$ respectively. Hence the names for involutions in Section 4 are indeed the same as the $\mathbb{ATLAS}$ class names, and we can now easily determine the class of any explicitly given involution.

Following [16], let $H=C_{\mathbb{B}}(d)\cong 2.{}^{2}E_{6}(2).2$ be a fixed $2A$ involution centralizer. This group has ten classes of involutions, whose labels in GAP [7] and the $\mathbb{ATLAS}$ are as follows:

[TABLE]

Since $H$ has shape $2.X.2$, it has an outer automorphism negating the outer classes, and there is an arbitrary choice of which is which of classes $\pm 2D$. For consistency with [16], we choose $+2D$ to be the class which fuses to $2A$ in the Baby Monster. On the other hand, the classes $\pm 2A$ are distinguished in their common centralizer $C=(2\times 2^{1+20}).\mathrm{U}_{6}(2).2$, in that the involution in the derived subgroup of $O_{2}(C)$ is in class $+2A$. It follows that $+2A$ fuses to $2B$ in the Baby Monster, while $-2A$ fuses to $2A$ in the Baby Monster. Indeed, computation using explicit matrices, and suitable class invariants as above, gives the full fusion of involutions from $H$ to ${\mathbb{B}}$. We find that GAP classes $2,4,175$ fuse to $2A$, and classes $3,5$ fuse to $2B$, while classes $176,177$ fuse to $2C$, and classes $6,7,178$ fuse to $2D$.

We will also need to know the fusion of involutions from the subgroups $\mathrm{Fi}_{23}$ and $\mathrm{HN}$. The easiest way to verify this is probably to use the words in [21] to find these subgroups explicitly, and compute a suitable class invariant as described above. We then see that classes $2A,2B,2C$ in $\mathrm{Fi}_{23}$ fuse to ${\mathbb{B}}$ classes $2A,2B,2D$ respectively, while classes $2A,2B$ in $\mathrm{HN}$ fuse to $2B,2D$ respectively in ${\mathbb{B}}$.

5.2. The permutation representation on $\{3,4\}$-transpositions

According to [16] the non-trivial suborbit lengths of ${\mathbb{B}}$ acting on the $13571955000$ cosets of $H$ are as follows:

• •

$3^{3}.5.7.13.17.19=3968055$, with point stabilizer $2.2^{1+20}.\mathrm{U}_{6}(2).2$;

• •

$2^{12}.3^{3}.11.19=23113728$, with point stabilizer $2^{2}\times F_{4}(2)$;

• •

$2^{20}.7.17.19=2370830336$, with point stabilizer $\mathrm{Fi}_{22}.2$; and

• •

$2^{8}.3^{3}.5.7.11.13.17.19=11174042880$, with point stabilizer $2^{1+20}.\mathrm{U}_{4}(3).2^{2}$.

We now compute the permutation characters of the action of $H$ on the first three of these suborbits. We use standard operations in GAP, using only the character tables of $H$ and certain of its subgroups. For simplicity we use the GAP labels for characters of $H$.

In the first case, the action on the suborbit is the permutation action of ${}^{2}E_{6}(2).2$ on the cosets of the $\mathrm{U}_{6}(2)$ maximal parabolic, and is known to have rank $5$ (see, for example, Theorem 4 in [17]). A straightforward combinatorial computation, using GAP, shows that the only way to get the character degrees adding to the correct number is for the degrees to be $1+1938+48620+1828332+2089164$. The trivial character is a constituent, because it is a permutation character, leaving $16$ possibilities for the signs on the other four constituents. It turns out that only one of these characters has non-negative values. This character is the sum of the irreducibles labelled $1,3,5,13,15$ in GAP.

In the second case, GAP computes possible class fusions from $F_{4}(2)$ into $H$, and we induce up the trivial character in each case. The answers are all the same. The permutation character is a subcharacter of this induced character, and it is easy to determine the character degrees, and then check all possibilities as above. The answer is the sum of irreducibles numbered $1,5,17,24$.

In the third case, similarly, we compute possible class fusions from $\mathrm{Fi}_{22}{:}2$ into $H$. There are then two possibilities for the induced trivial character, and they differ by multiplying the outer elements of $\mathrm{Fi}_{22}{:}2$ by the central involution of $H$. But we know that in the point stabilizer $\mathrm{Fi}_{22}{:}2$ the $2D$ involutions fuse to $2A$ in ${\mathbb{B}}$ (if necessary we can verify this computationally using the subgroup $S_{3}\times\mathrm{Fi}_{22}{:}2$ in our certified copy of the Baby Monster), which distinguishes the two cases. The answer is the sum of characters numbered $1,3,5,13,17,28,49,76,190,192,196,202,210,217$.

It is not necessary to compute the full permutation character of ${\mathbb{B}}$ on the cosets of $H$, which would involve computing the fourth suborbit case as well. Later on we will however need to compute the values on a few selected classes.

5.3. Fusion of $3$-elements

Computationally, using a certified copy of the Baby Monster, and words provided in [21], we find two subgroups $S_{3}\times\mathrm{Fi}_{22}{:}2$ and $3^{1+8}{:}2^{1+6}.\mathrm{U}_{4}(2).2$, which normalize cyclic subgroups of order $3$. The corresponding elements of order $3$ can be distinguished by the trace in the $4370$ dimensional representation mod $2$, so do not fuse in ${\mathbb{B}}$. We use the $\mathbb{ATLAS}$ labels $3A$ and $3B$ for these two conjugacy classes.

Conversely, note that $\mathrm{Fi}_{23}$ contains a Sylow $3$-subgroup of ${\mathbb{B}}$ so every $3$-element in ${\mathbb{B}}$ is conjugate to an element of $\mathrm{Fi}_{23}$. Moreover, we know the fusion from $2.\mathrm{Fi}_{22}$ to $\mathrm{Fi}_{23}$, and in particular, every $3$-class in $\mathrm{Fi}_{23}$ is represented in $2.\mathrm{Fi}_{22}$ and therefore in $H$. Using the fact that $H$-classes $-2A$ and $+2D$ are in $2A$, and computing structure constants in $H$, we get that $H$-classes $3A$ and $3B$ fuse in ${\mathbb{B}}$. Hence there are exactly two classes of elements of order $3$ in ${\mathbb{B}}$.

We now show that a $3A$-element $x$ has centralizer $C_{{\mathbb{B}}}(x)\cong 3\times\mathrm{Fi}_{22}{:}2$ in ${\mathbb{B}}$. We know its centralizer is at least that (either computationally, as above, or see [16]). On the other hand, the number of ${\mathbb{B}}$-conjugates of $x$ is at least one-third of the product of the length of the whole orbit with the length of the relevant suborbit. This number $13571955000\times 2370830336/3$ is equal to the index of $3\times\mathrm{Fi}_{22}{:}2$ in ${\mathbb{B}}$, and the claim is proved.

Next we show that the subgroup $3^{1+8}.2^{1+6}.\mathrm{U}_{4}(2)$ computed above is the full centralizer of a $3B$-element. To do this we need to know the value on $3B$ of the full permutation character of ${\mathbb{B}}/H$. Equivalently, the value of the permutation character of the last orbit above on $H$-class $3C$. Recall that the point stabilizer in the last orbit is $H_{5}=2^{1+20}.\mathrm{U}_{4}(3).2^{2}$. We use the GAP function PossibleClassFusions applied to the character tables of $\mathrm{U}_{4}(3)$ and $2.\mathrm{U}_{4}(3)$ to get the fusion of $3$-elements from $H_{5}$ to $H$. The result is that $\mathbb{ATLAS}$ class $3A$ in $\mathrm{U}_{4}(3)$ fuses to $3C$ in $H$, while all other classes of elements of order $3$ fuse to $3A$ or $3B$ in $H$. Hence the value on $H$-class $3C$ of the permutation character of $H$ on this orbit is $|C_{H}(3C)|/|C_{H_{5}}(3A)|=2^{11}.3^{9}/2^{8}.3^{6}=2^{3}.3^{3}=216$. Therefore the value of the whole permutation character of $B/H$ on class $3C$ is $1620$.

Hence we know $|C_{\mathbb{B}}(3B)|/|C_{H}(3C)|=1620=2^{2}.3^{4}.5$, and the $3C$-centralizer in $H$ has order $2^{11}.3^{9}$, so we deduce the order of the $3B$-centralizer in ${\mathbb{B}}$ is $2^{13}.3^{13}.5$ and the claim follows.

5.4.

Elements of orders $7,11,13,17,19$ and $23$

From [16] (Lemma 6.11) we get $|N_{\mathbb{B}}(7^{2})|=2^{6}.3^{2}.7^{2}$ and $|C_{\mathbb{B}}(7^{2})|=2^{2}.7^{2}$. The only subgroup of $\mathrm{GL}_{2}(7)$ of order $2^{4}.3^{2}$ is $3\times 2S_{4}$, which is transitive on non-zero elements of $7^{2}$. Hence there is a single class of elements of order $7$. (This can also be verified computationally in a certified copy of the Baby Monster.)

In $H$ we have a $7A$-centralizer $7\times 2.\mathrm{L}_{3}(4).2$. To show that the centralizer in ${\mathbb{B}}$ is no bigger, we follow the same strategy as for $3B$ elements above, although it is slightly more complicated since both classes $7A$ and $7B$ fuse to $7A$ in ${\mathbb{B}}$. The $7$-elements in $H_{5}$ fuse to class $7B$ in $H$. Hence the value of the permutation character of ${\mathbb{B}}/H$ on $H$-class $7A$ is $121$. (As a check, $|C_{H}(7B)|=4704=2^{5}.3.7^{2}$ and $|C_{H_{5}}(7B)|=2^{4}.7$, so the character value of the last orbit on $H$-class $7B$ is $2.3.7=42$. This implies the value of the permutation character of ${\mathbb{B}}/H$ on $H$-class $7B$ is also $121$, as it must be.) Therefore $|C_{\mathbb{B}}(7A)|/|C_{H}(7A)|+|C_{\mathbb{B}}(7A)|/|C_{H}(7B)|=121$, so that $|C_{\mathbb{B}}(7A)|=2^{8}.3^{2}.5.7$.

For the remaining primes in the list, $11$, $13$, $17$, $19$ and $23$, most of the information we need is already in [16]. Lemma 6.13 of [16] says the centralizer order of an element of order $17$ is $2^{2}.17$, and the normalizer has order $2^{6}.17$, so there is a single class of elements of order $17$. Lemma 6.8 of [16] says that the order of the Sylow $11$-normalizer is $2^{4}.3.5^{2}.11$, and the centralizer of an element of order $11$ is $S_{5}\times 11$. Hence the normalizer is $S_{5}\times 11{:}10$. In Lemma 6.12 of [16] there are two possibilities for the normalizer of an element of order $13$. But the normalizer of such an element in $F_{4}(2)$ is just $13{:}12$, so from the proof of Lemma 6.12 we get that the $13$-centralizer in ${\mathbb{B}}$ is $13\times S_{4}$, and the normalizer is $13{:}12\times S_{4}$.

The Sylow $19$-subgroup is self-centralizing in $H/\langle d\rangle$, so the Sylow $2$-subgroup of $C_{\mathbb{B}}(19)$ is of order $2$, containing a $2A$-element. Since $|\mathrm{Fi}_{22}|$ is not divisible by $19$, that forces $N_{\mathbb{B}}(19)$ to lie in $H$. Lemma 6.20 of [16] says that the $23$-normalizer contains $2\times 23{:}11$, and that the Sylow $2$-subgroup of the $23$-normalizer has order $2$; we know (from Lemmas 7.13, 7.14, 7.15 and 7.17 of [16]) that all Sylow subgroups of the normalizer are cyclic. From the discussion earlier in this section, we know that the normalizer does not contain elements of order $7,13,17$, or $19$. The normalizing $11$ rules out $47$ and $31$, by the Frattini argument. This leaves $3,5$. We know the $3$-centralizers, so $3$ is ruled out. Finally $5$ is ruled out because $|{\mathbb{B}}|/(2.5.11.23)\not\equiv 1\pmod{23}$.

5.5. The elements of order $5$

The subgroup $\mathrm{HN}$ (constructed explicitly in our certified copy of ${\mathbb{B}}$) contains a full Sylow $5$-subgroup. Every element of order $5$ in $\mathrm{HN}$ centralizes an involution, which we know fuses to $2B$ or $2D$ in ${\mathbb{B}}$. Moreover, every element of order $5$ in $C_{\mathbb{B}}(2B)$ or $C_{\mathbb{B}}(2D)$ centralizes an element of order $3$. But $C_{\mathbb{B}}(3A)$ and $C_{\mathbb{B}}(3B)$ contain just one class of elements of order $5$ each, so there are at most two classes of elements of order $5$ in ${\mathbb{B}}$. On the other hand, we find two classes of $5$-elements with different traces. Hence there are exactly two classes.

The usual argument gives the order of $C_{\mathbb{B}}(5A)$. We have $|C_{H}(5A)|=2^{8}.3^{2}.5^{2}.7$ and $|C_{H_{5}}(5A)|=2^{7}.5$ so the value of the permutation character of the last orbit on this class is $2.3^{2}.5.7=630$. Hence the full permutation character has value $630+470=1100$ on ${\mathbb{B}}$ class $5A$. Therefore $|C_{\mathbb{B}}(5A)|=1100|C_{H}(5A)|=2^{10}.3^{2}.5^{4}.7.11$, which is the order of $5\times\mathrm{HS}{:}2$. Hence the $5A$-normalizer is $5{:}4\times\mathrm{HS}{:}2$.

Computationally, using the matrices and words provided in [21], we find a subgroup $5^{1+4}.2^{1+4}.A_{5}.4$, normalizing a cyclic group of order $5$, which must therefore be of $5B$ type. We shall show that the normalizer is no bigger than this. We know that there is no $5B$ element in the centralizer of any element of order $7,11,13,17,19,23$, or of a $3A$. Also $47$ and $31$ are $3\mod 4$ and do not centralize an involution, so do not centralize a $5B$ by the Frattini argument. Hence the centralizer of a $5B$ is a $\{2,3,5\}$-group, and contains the full Sylow $5$-subgroup of ${\mathbb{B}}$, so only the Sylow $2$- or $3$-subgroup could grow.

Now the centralizer of a $5$-element in $C_{\mathbb{B}}(3B)=3^{1+8}.2^{1+6}.\mathrm{U}_{4}(2)$ is just a cyclic group of order $30$, so the Sylow $3$-subgroup of the $5B$-centralizer has order $3$. Since $C_{\mathbb{B}}(2A)$ and $C_{\mathbb{B}}(2C)$ contain no $5B$, we look in $C_{\mathbb{B}}(2B)$ and $C_{\mathbb{B}}(2D)$. In $C_{\mathbb{B}}(2B)$ only the $5A$ class of $\mathrm{Co}_{2}$ fuses to ${\mathbb{B}}$ class $5B$, and we see the centralizer $(2^{1+2}\times 5^{1+2}).2A_{4}$ of order $2^{6}.3.5^{3}$. In $C_{\mathbb{B}}(2D)$ we see centralizer order $2^{7}.3.5^{2}$. In neither case does the Sylow $2$-subgroup grow. Thus we know the orders of all the Sylow subgroups of $C_{\mathbb{B}}(5B)$, and therefore the order of $C_{\mathbb{B}}(5B)$.

5.6. Primes $47$ and $31$

The order of the $47$-normalizer now divides $47.2.23.31$ and Sylow’s theorem implies it is $47.23$. Finally the order of the $31$-normalizer divides $31.30$, so is $31.15$ by Sylow’s Theorem.

6. Obtaining the class list

Our strategy for obtaining the list of conjugacy classes in the Baby Monster is first to determine the classes of even order elements, by computing the character tables of subgroups containing the four distinct involution centralizers, and noting down the conjugacy classes of elements in each subgroup that power to the relevant involution class. (The centralizers of involutions in classes $2A$, $2B$, $2C$ are in fact maximal, although it is not necessary to know this, so we have no choice but to use the involution centralizer itself in these cases.) At the same time, we note down the length of each such class. A similar computation for odd-order elements in the centralizers of elements of odd prime order is trivial in comparison.

In fact, there is a great deal of redundancy in the information that we have computed, and classes of elements whose order is divisible by two primes can be computed in two different ways. This provides a robust check on these results, in particular for the large number of classes of elements of order divisible both by $2$ and by an odd prime.

6.1. Involution centralizers in the Baby Monster

The character table of the $2A$-centralizer is known by [5] and the computations shown in [3]. The $2C$-centralizer has the structure $(2^{2}\times F_{4}(2)).2<D_{8}\times F_{4}(2).2$, and its character table is determined by those of the subgroups $2^{2}$ and $F_{4}(2)$ and the factor groups $D_{8}$ and $F_{4}(2).2$, hence it is known.

The $2D$-centralizer is contained in subgroups of the structure $2^{(8+1)+16}.\mathrm{S}_{8}(2)$ in ${\mathbb{B}}$. Such subgroups can be constructed explicitly in a certified copy of ${\mathbb{B}}$, using the straight line program from [21]. The character table of this subgroup can be verified by restricting the $2$-modular degree $4370$ representation of ${\mathbb{B}}$ to the subgroup, finding a faithful $180$-dimensional subquotient of this module, and computing the character table from this matrix representation using the MAGMA computer algebra system [2]. (This had been done by E. O’Brien in 2007, but we repeat the computations in order to make sure that only explicitly verified data are used.) In particular, this verification includes a verification that the given subgroup contains the full $2D$-centralizer.

The $2B$-centralizer has the structure $2^{1+22}.\mathrm{Co}_{2}$. The character table has been computed in [12] but the arguments assume the character table of ${\mathbb{B}}$. In the remainder of this section, we describe briefly how we verify this character table. Full details can be found in [4].

First we restrict the certified $3$- and $5$-modular representations of degree $4371$ of ${\mathbb{B}}$ to the $2B$-centralizer, using the straight line program from [21]; the composition factors of the module have the dimensions $23$, $2300$, and $2048$ in both cases. Next, we find an orbit of length $4600$ in the $2300$-dimensional module over the field of order $3$. The action on this orbit yields a faithful permutation representation of the factor group $2^{22}.\mathrm{Co}_{2}$. We compute class representatives for this factor group, and let MAGMA compute its character table.

The $2048$-dimensional module is faithful. We compute the class fusion under the epimorphism from $2^{1+22}.\mathrm{Co}_{2}$ to $2^{22}.\mathrm{Co}_{2}$, and the Brauer characters of our $3$- and $5$-modular representations for this module. Now $2^{1+22}$ has a unique faithful irreducible representation in every characteristic except $2$, and this representation has dimension $2^{11}=2048$, and extends uniquely to $2^{1+22}.\mathrm{Co}_{2}$. If $\chi$ denotes the character of the ordinary representation of $2^{1+22}.\mathrm{Co}_{2}$ obtained in this way, then all faithful irreducible characters of $2^{1+22}.\mathrm{Co}_{2}$ arise as tensor products of $\chi$ with the irreducibles of the factor group $\mathrm{Co}_{2}$. In particular the Brauer characters computed above lift to $\chi$, and therefore we obtain the values of $\chi$ on all classes of elements whose order is not divisible by $15$. For details of how the remaining values were obtained, see [4].

Once the character tables of (overgroups of) all the involution centralizers are available, we can read off from these tables all the conjugacy classes of elements that power to each of the involutions, together with the centralizer orders. This gives us a complete list of all conjugacy classes of even-order elements.

6.2. Elements of odd order

For primes $p\geq 11$ it is now almost a triviality to write down the classes of elements of odd order divisible by $p$. For $p=7$, we have that $N_{\mathbb{B}}(7A)$ is contained in $H$, so the relevant classes can be read off from the classes of elements of $H$ that power into class $7A$. (Note that the $7A$-centralizer in $H$ has the shape $7\times 2.\mathrm{L}_{3}(4){:}2_{2}$, and the $2_{2}$ automorphism swaps the $\mathrm{L}_{3}(4)$-classes $5A$ with $5B$.)

For elements powering into $5A$, we read off the classes and their centralizer orders from the $\mathbb{ATLAS}$ character table of $\mathrm{HS}{:}2$. Similarly, for elements powering into $3A$, use the table for $\mathrm{Fi}_{22}{:}2$, but note that there are some classes missing in the $\mathbb{ATLAS}$ character table for $\mathrm{Fi}_{22}{:}2$: these only affect the calculations for elements of order $30$, which have already been dealt with in the $5A$-centralizer.

In the cases $3B$ and $5B$ again, GAP contains character tables of the respective normalizers. However, it is not recorded exactly what information was used to calculate these tables. Therefore we re-calculate them (see [4]). In conclusion, we find that the list of odd-order elements and their centralizer orders agrees with the $\mathbb{ATLAS}$.

7. Computing the irreducible characters of the Baby Monster

From the previous sections, we know that ${\mathbb{B}}$ contains subgroups of the structures $2.{}^{2}E_{6}(2).2$, $\mathrm{Fi}_{23}$, and $\mathrm{HN}.2$. The ordinary character tables of these groups have been verified (see [5]) and thus may be used in our computations. The class fusions from these subgroups to ${\mathbb{B}}$ can be computed with the methods available in GAP [7]. Moreover, in Section 6.1, we have computed the character table of the $2B$-centralizer in ${\mathbb{B}}$. The class fusion from $2^{1+22}.\mathrm{Co}_{2}$ to ${\mathbb{B}}$ is determined by evaluating the three representations of ${\mathbb{B}}$ at the class representatives of $2^{1+22}.\mathrm{Co}_{2}$, and applying the invariants from Section 4.

Thus we can induce the irreducible characters from these subgroups to ${\mathbb{B}}$. Using the power maps of ${\mathbb{B}}$, we induce also the irreducibles of all cyclic subgroups of ${\mathbb{B}}$. Now we proceed in two steps.

In the first step, we assume that ${\mathbb{B}}$ has an ordinary irreducible representation $\chi$ of degree $4371$ such that the reductions modulo $3$ and $5$ are (irreducible and) equivalent to the representations we have used in the previous sections, and such that the reduction modulo $2$ has one trivial composition factor and one that is equivalent to the representation we have used above. Then the Brauer character values of our representations yield the values of $\chi$, except on the classes of elements with order divisible by $30$, and the missing values are uniquely determined by the obvious bounds. If we add $\chi$ and the trivial character of ${\mathbb{B}}$ to the list of induced characters then applying standard character-theoretic techniques such as LLL reduction yields a complete list of irreducible characters for ${\mathbb{B}}$, which coincides with the characters in the $\mathbb{ATLAS}$ table of ${\mathbb{B}}$.

In the second step, we do not want to assume the existence of the ordinary character $\chi$, and try to apply the character-theoretic criteria to the safe list of induced characters. This way, we do not get any irreducible character. However, we can show that $30$ class functions from the list of irreducibles computed in the first step lie in the lattice spanned by the induced characters. Thus these class functions are verified as irreducible characters of ${\mathbb{B}}$. Now we form symmetrizations and tensor products of the known irreducible characters, and the lattice spanned by the known characters of ${\mathbb{B}}$ contains all the missing irreducibles computed in the first step. Thus we are done. Again, the details of these constructions can be found in [4].

Acknowledgements

We thank Chris Parker for significant contributions to the original version of this paper, and we thank the referee for helpful comments that enabled us to avoid the need for them.