Primitive almost simple IBIS groups with sporadic socle

In all of these cases, we are able to construct the permutation representation of $G$ acting on cosets of $H$ in Magma by first constructing $G$ as a matrix group, finding its maximal subgroups and using the LMGCosetImage function. We prove most groups are not IBIS by applying (T1). The remaining cases are when $(G,H,b(G))$ is one of $(\operatorname{M}_{11},A_{6}.2,4)$, $(\operatorname{M}_{12},\operatorname{M}_{11},5)$, $(\operatorname{M}_{22},\text{L}_{3}(4),5)$, $(\operatorname{M}_{23},\operatorname{M}_{22},6)$ or $(\operatorname{M}_{24},\operatorname{M}_{23},7)$; we prove they are IBIS by enumerating a set of orbit representatives on $b(G)$-tuples, and showing that each is either an irredundant base, or is not a base.

Here at least one of the permutation representations equivalent to $G$ acting on cosets of $H$ is not readily available. In most cases however, we are able to construct $H$ using the AtlasSubgroup command as part of the AtlasRep package of GAP [2, 6]. We then apply (T2) in each case and deduce that $G$ is not IBIS.

In view of Proposition 1.2, we have $H\in\{G_{2}(5),3.\text{McL}:2\}$, where $b(G)=3$ in each case.

Let $H=G_{2}(5)$. We appeal to (T3) by considering the centraliser $K=C_{G}(g_{5})$ of an element $g_{5}\in H$ of order 5. We find $H$ has two classes of elements of order 5 and, by the centraliser orders, we deduce that both classes are contained in the class labelled 5A in $G$. Now $K=C_{G}(g_{5})$ is maximal in $G$ [5, p. 174] and, replacing $g_{5}$ with a conjugate if necessary, a construction of $K$ is available as a straight line program in the Online Atlas [13]. Without loss, we choose $g_{5}$ to be in class 5A in $H$, so that $|C_{H}(g_{5})|=2^{3}.3.5^{6}$, and find candidates for $C_{H}(g_{5})<K$ using the Subgroups command in Magma. In order to construct an partial irredundant base for $K$ of length 3, we first fix the trivial coset $H$ so that we now must find a partial irredundant base for $K_{2}=C_{H}(g_{5})$ of size 2. We then randomly search $K$ for two elements $x_{1},x_{2}$ such that the following conditions hold:

1. (1)

   $x_{1}x_{2}^{-1}\notin H$, so that $Hx_{1}$ and $Hx_{2}$ are distinct.

2. (2)

   $K_{2}>K_{2}\cap K_{2}^{x_{1}}>1$, and

3. (3)

   $K_{2}\cap K_{2}^{x_{1}}>K_{2}\cap K_{2}^{x_{1}}\cap K_{2}^{x_{2}}>1$ so that $\{H,Hx_{1},Hx_{2}\}$ forms a partial irredundant base for $K$ by (T2).

Choosing $K$, $K_{2}$ in this way allows us to verify each of these conditions in $K$ rather than $G$, which significantly reduces the computational resources required. We find an appropriate $x_{1},x_{2}$ via random search and therefore prove that $G$ is not IBIS.

If instead $H=3.\text{McL}:2$, we repeat the same process, instead considering the centraliser of an involution in class 2A of $H$. We deduce that $G$ is not IBIS.

By Proposition 1.2, we only need consider the action of $G$ on right cosets of one of its maximal subgroups $H\in\{2^{11}:\operatorname{M}_{24},2^{1+12}:3.\operatorname{M}_{22}.2,2^{10}:\mathrm{L}_{5}(2)\}$, where $b(G)=3$ in each case.

Suppose $H=2^{11}:\operatorname{M}_{24}$. Now $H$ is the stabiliser of a vector in the irreducible 112-dimensional representation of $G$ over $\mathbb{F}_{2}$ [5, p.190]. Generators for $G$ and $H$ in this representation, along with a vector $v$ fixed by $H$, are given in the Online Atlas [13]. The action of $G$ on cosets of $H$ is then permutationally isomorphic to the action of $G$ on $v^{G}$. We consider a subgroup $K\cong\operatorname{M}_{24}$ of $H$ and find that $v$ along with two further vectors in $v^{G}$ obtained by random search have stabiliser order chain $[|M_{24}|,16,2]$ in $K$. The result follows applying (T3).

Now suppose $H=2^{1+12}:3.\operatorname{M}_{22}.2$. We employ technique (T3) with the centraliser of an element $g_{5}\in H$ of order 5. There is a single class of elements of order 5 in $G$ and so we have $|C_{G}(g_{5})|=2^{6}.3.5.7$, while $|C_{H}(g_{5})|=2^{6}.3.5$. Two elements $x_{1},x_{2}\in C_{G}(g_{5})$ are representatives for the same coset in $G/H$ if and only if $x_{1}x_{2}^{-1}\in C_{H}(g_{5})$. We find appropriate $x_{1},x_{2}$ via random search such that by (T2) we can deduce that $H,Hx_{1}$ and $Hx_{2}$ form a partial irredundant base for $C_{G}(g_{5})$, so $G$ is not IBIS.

Finally, suppose $H=2^{10}:\mathrm{L}_{5}(2)$. We apply (T3) with a subgroup $K\cong L_{5}(2)$ of $H$. The construction of such a subgroup $K$ is available in the Online Atlas, and we find that $K$ fixes a non-zero vector $w\in V$. As before, we consider the action of $G$ on $w^{G}$. By random search, we find $h_{1},h_{2}\in G$ such that $H=G_{w}>G_{w,w^{h_{1}}}>G_{w,w^{h_{1}},w^{h_{2}}}>1$, so $G$ is not IBIS.

By Lemma 1.1 and Proposition 1.2, if $G$ is IBIS, then $H\in\{{}^{3}\mathrm{D}_{4}(2):3,2^{5}.\mathrm{L}_{5}(2)\}$ and $b(G)=3$. In each case, straight line programs for the generators of $H$ in terms of standard generators of $G$ are available in the Online Atlas. We proceed by applying (T3) in the same fashion as in Section 1.3. More explicitly, if $H={}^{3}\mathrm{D}_{4}(2):3$, then we apply (T3) by considering the centraliser of an involution $g_{2}$ lying in the class labelled 2A of $H$. If instead $H=2^{5}.\mathrm{L}_{5}(2)$, we again consider the centraliser of a 2A involution in $H$.

By Proposition 1.2, it remains to check that the action of $G$ on cosets of $H=2.\text{B}$, where $b(G)=3$. First fix the identity coset, whose stabiliser is $H$. If $G$ is IBIS, then every irredundant base of $H$ on $G/H$ has size 2. Let $g_{17}$ be an element of order 17. We consider $K=C_{H}(g_{17})$ with the view of employing (T3). We may construct $C=|C_{G}(g_{17})|$ in GAP as a subgroup of its maximal overgroup $(L_{3}(2)\times S_{4}(2):2).2<G$ using the Online Atlas [5, p.234]. We know from the character table of $H$ [5, p.210–219] that $|K|=136$. Now $C$ has a single conjugacy class of subgroups of order 136, so without loss of generality, we may fix one of them and set it to be $K$. Now $C$ is divisible by 3 and 7, while $|K|$ is not, so take $g_{3},g_{7}\in C_{G}(g_{17})$ to be elements of orders 3 and 7 respectively. Clearly $H\neq Hg_{3},Hg_{7}$, and we must also verify that $g_{3}g_{7}^{-1}\notin H$. We observe that $k\in K$ fixes $Hg_{3}$ if and only if $k^{g_{3}}\in H$. Since $k^{g_{3}}\in C$, we must have $k^{g_{3}}\in H\cap C=K$, and a similar argument holds for $Hg_{7}$. Therefore, we check directly in our construction of $C$ and $K$ in GAP that there exist $g_{3}$, $g_{7}$ such that: (1) $g_{3}g_{7}^{-1}\notin H$ so that $Hg_{3}$, $Hg_{7}$ are distinct, (2) $K_{1}=K^{g_{3}}\cap K<K$ and (3) $1<K^{g_{7}}\cap K_{1}<K_{1}$. Together, these demonstrate that $K$ has a partial irredundant base of size 2 on $G/H$, hence so does $H$. Therefore, $G$ is not IBIS.

By Proposition 1.2, we need to consider

$$H\in\{2.{}^{2}.E_{6}(2).2,2^{1+22}.\mathrm{Co}_{2},\mathrm{Fi}_{23},2^{9+16}.\mathrm{PSp}_{8}(2),\rm{Th},(2^{2}\times F_{4}(2)):2,2^{2+10+20}.(M_{22}:2\times S_{3})\}$$

The action of $G$ on cosets of $H$ has base size 3 in all cases except $H=2.{}^{2}.E_{6}(2).2$, where $b(G)=4$.

First let $H=2.{}^{2}.E_{6}(2).2$. We can construct $H$ as the centraliser of an element $t_{0}$ in conjugacy class $2A$ of $G$. The orbit lengths of $H$ on 2A involutions $t\in B$ are given in [5, p. 216] and we reproduce them in Table 1.

Let $t_{i}$ be an orbit representative of the orbit labelled $i$ in Table 1. We claim that $\{t_{0},t_{1},t_{2},t_{3}\}$ forms a partial irredundant base for $G$, implying $G$ is not IBIS. Clearly $G_{t_{0}}=H$ by definition and $|G_{t_{0},t_{1}}|=|C_{H}(t_{1})|=2^{38}.3^{6}.5.7.11$ by Table 1. Choosing $t_{2}$ and $t_{3}$ appropriately, we find that the Sylow 2-subgroup of $|G_{t_{0},t_{1},t_{2}}|$ has order $2^{26}$, while the Sylow 2-subgroup of $|G_{t_{0},t_{1},t_{2},t_{3}}|$ has order $2^{18}$. Therefore, these groups are distinct and non-trivial, so the claim is proved.

Now let $H=2^{1+22}.\mathrm{Co}_{2}$. Similar to the last case, $H$ can be constructed as the centraliser of an involution $t_{0}$ in class 2B of $B$, and we can instead consider the action of $G$ on its class 2B. Müller [10, Table 1] computed the orbit lengths of $H$ on 2B involutions. The smallest non-trivial orbit has length 93150 and so the size of a stabiliser $K$ in $H$ of an element $t_{1}$ of this orbit is larger than the number of cosets of $H$ in $G$. Therefore, it is impossible to construct an irredundant base of size 3 that begins with $\{t_{0},t_{1}\}$, so $G$ is not IBIS.

Now let $H=\rm{Fi}_{23}$. The $H$-orbits on cosets of $H$ were computed by Müller et al. [11, Table 2]. The Sylow 7-subgroup of $H$ has order 7, as do the stabilisers $K_{2}\cong O_{8}^{+}(3):2_{2}$ and $K_{3}\cong\mathrm{PSp}_{8}(2)$ of elements in the non-trivial orbits labelled 2 and 3 in [11, Table 2] respectively. Hence, taking appropriate representatives $Hx_{2},Hx_{3}$ of these orbits, we achieve an irredundant partial base $\{H,Hx_{2},Hx_{3}\}$ of size 3 stabilised by an element of order 7. Since the stabiliser is not trivial, $G$ is not IBIS.

Next suppose $H=2^{2+10+20}(M_{22}:2\times S_{3})$. Neunhöffer et al. [12, Table 1] computed a subset of the $H$-suborbits of $B$ acting on cosets of $H$. Although only the orbit lengths and sizes of the corresponding stabilisers in $H$ are available, this is sufficient to show $G$ is not IBIS. There is a single class of elements of order 5 in $H$, and two distinct orbits of $H$ on cosets with stabiliser sizes 240 and 120. Therefore, by choosing representatives of these orbits appropriately, we can construct a partial irredundant base of size three stabilised by an element of $G$ of order 5. Therefore, $G$ is not IBIS.

For the remaining maximal subgroups $H_{4}=2^{9+16}.\mathrm{PSp}_{8}(2)$, $H_{5}=\rm{Th}$ and $H_{6}=(2^{2}\times F_{4}(2)):2$ we adopt a different approach. In the following construction, we will use $H$ to refer to any one of the groups $H_{4},H_{5}$ or $H_{6}$. Let $g_{6}\in H$ have order 6, and let $g_{2}=g_{6}^{3}$. We will construct distinct cosets $Hx$ and $Hy$ such that $Hx$ is fixed by $g_{2}$ and $g_{6}$, while $Hy$ is fixed by $g_{2}$, but not $g_{6}$. This will show that $\{H,Hx,Hy\}$ form a partial irredundant base for $G$ in its coset action on $H$, and so $G$ is not IBIS.

Now, $g$ fixes $Hx$ if and only if $g^{x^{-1}}=xgx^{-1}\in H$, and similarly for $Hy$. Hence, it is sufficient to exhibit $x$ and $y$ such that $g_{6}^{x^{-1}},g_{2}^{x^{-1}},g_{2}^{y^{-1}}\in H$, while $g_{6}^{y^{-1}}\notin H$. Notice that we may simplify the problem slightly by taking $x,y\in HC_{G}(g_{2})$. We must then check that $g_{6}^{x^{-1}}\in H$ $g_{6}^{y^{-1}}\notin H$ and that $x,y,xy^{-1}\notin H$ to ensure that $H$, $Hx$ and $Hy$ are distinct. Since $HC_{G}(g_{6})\subseteq HC_{G}(g_{2})$, the probability of finding $x\in C_{G}(g_{2})$ such that $g_{6}^{x^{-1}}\in H$ is proportional to

$$\frac{|C_{G}(g_{6})|}{|C_{G}(g_{2})|}=\frac{|C_{G}(g_{6})|{|C_{H}(g_{2})|}}{|C_{G}(g_{2})|{|C_{H}(g_{6})|}}.$$

We try to maximise this quantity in our choice of the class of $g_{6}$, and using the Character Table Library in GAP (including the labelling of classes given there), we choose $g_{6}$ from classes $6C$, $6B$ and $6Q$ in $H_{4}$, $H_{5}$ and $H_{6}$ respectively.

We now turn to a computational search in Magma. The relevant code to reproduce the procedure we describe is available on the author’s website. We are able to construct $G$ as a matrix group in a 4370-dimensional representation over $\mathbb{F}_{2}$, and construct each of the maximal subgroups $H$ using straight-line programs on the standard generators, which are available in [13]. We observe that each maximal subgroup acts reducibly on the underlying vector space $V$, so we compute the submodules preserved by each maximal subgroup. In each case, we are able to find a relatively small submodule on which $H$ acts faithfully. We project onto this submodule so that we are able to find the conjugacy classes of $H$ using the LMGClasses function. Once we have found an appropriate element $g_{6}$, we then find the preimage in the 4370-dimensional representation of $H$, and define $g_{2}=g_{6}^{3}$. We compute $C_{G}(g_{2})$ using CentraliserOfInvolution function, and search for $x$ and $y$ satisfying the above conditions by random selection. Since determining membership of a random element of $G$ in $H$ is very time-consuming, we instead equivalently check whether the element preserves a submodule fixed by $H$. We are able to find an appropriate $x$ and $y$ in each case, so infer that $G$ is not IBIS in its coset actions on $H_{4}$, $H_{5}$ and $H_{6}$.

The proof of Theorem 1 is now complete.