The intersection graph of a finite simple group has diameter at most 5

Abstract

Let G be a non-abelian finite simple group. In addition, let $\Delta _G$ be the intersection graph of G, whose vertices are the proper non-trivial subgroups of G, with distinct subgroups joined by an edge if and only if they intersect non-trivially. We prove that the diameter of $\Delta _G$ has a tight upper bound of 5, thereby resolving a question posed by Shen (Czechoslov Math J 60(4):945–950, 2010). Furthermore, a diameter of 5 is achieved only by the baby monster group and certain unitary groups of odd prime dimension.

1 Introduction

For a finite group G, let $\Delta _G$ be the intersection graph of G. This is the graph whose vertices are the proper non-trivial subgroups of G, with two distinct vertices $S_1$ and $S_2$ joined by an edge if and only if $S_1 \cap S_2 \ne 1$. We write $d(S_1,S_2)$ to denote the distance in $\Delta _G$ between vertices $S_1$ and $S_2$, and if these vertices are joined by an edge, then we write $S_1 \sim S_2$. Additionally, $\mathrm {diam}(\Delta _G)$ denotes the diameter of $\Delta _G$.

Csákány and Pollák [5] introduced the graph $\Delta _G$ in 1969 as an analogue of the intersection graph of a semigroup defined by Bosák [1] in 1964. For finite non-simple groups G, Csákány and Pollák determined the cases where $\Delta _G$ is connected, and proved that, in these cases, $\mathrm {diam}(\Delta _G) \leqslant 4$ (see also [14, Lemma 5]). It is not known if there exists a finite non-simple group G with $\mathrm {diam}(\Delta _G) = 4$.

Suppose now that G is a non-abelian finite simple group. In 2010, Shen [14] proved that $\Delta _G$ is connected, and asked two questions: does $\mathrm {diam}(\Delta _G)$ have an upper bound? If yes, does the upper bound of 4 from the non-simple case also apply here? In the same year, Herzog, Longobardi, and Maj [7] independently showed that the subgraph of $\Delta _G$ induced by the maximal subgroups of G is connected with diameter at most 62. As each proper non-trivial subgroup of G is adjacent in $\Delta _G$ to some maximal subgroup, this implies an upper bound of 64 for $\mathrm {diam}(\Delta _G)$, resolving Shen’s first question. Ma [12] reduced this upper bound to 28 in 2016. In the other direction, Shahsavari and Khosravi [13, Theorem 3.7] proved in 2017 that $\mathrm {diam}(\Delta _G) \geqslant 3$.

In this paper, we significantly reduce the previously known upper bound of 28 for $\mathrm {diam}(\Delta _G)$, and show that the new bound is best possible. In particular, we prove the following theorem, which resolves Shen’s second question with a negative answer.

Theorem 1.1

Let G be a non-abelian finite simple group.

• (i) $\Delta _G$ is connected with diameter at most 5.

• (ii) If G is the baby monster group ${\mathbb {B}}$, then $\mathrm {diam}(\Delta _G) = 5$.

• (iii) If $\mathrm {diam}(\Delta _G) = 5$ and $G \not \cong {\mathbb {B}}$, then G is a unitary group $\mathrm {U}_n(q)$, with n an odd prime and q a prime power.

Remark 1.2

Using information from the Atlas [4], we can show that if $S_1$ and $S_2$ are vertices of $\Delta _{{\mathbb {B}}}$ with $d(S_1,S_2) = 5$, then $|S_1| = |S_2| = 47$.

Remark 1.3

If $G \in \{\mathrm {U}_3(3), \mathrm {U}_3(5),\mathrm {U}_5(2)\}$, then G has no maximal subgroup of odd order [11, Theorem 2]. As we will explain in the proof of Theorem 1.1, this implies that $\mathrm {diam}(\Delta _G) \leqslant 4$. Indeed, we can use information from the Atlas [4] to show that $\mathrm {diam}(\Delta _{\mathrm {U}_3(3)}) = 3$. Furthermore, even though $\mathrm {U}_3(7)$ has a maximal subgroup of odd order, we deduce from calculations in Magma [2] that $\mathrm {diam}(\Delta _{\mathrm {U}_3(7)}) = 4$. On the other hand, we can adapt the proof of Theorem 1.1(ii), with the aid of several Magma calculations, to show that $\mathrm {diam}(\Delta _{\mathrm {U}_7(2)}) = 5$.

It is an open problem to classify the finite simple unitary groups G with $\mathrm {diam}(\Delta _G) = 5$.

2 Proof of Theorem 1.1

In order to prove Theorem 1.1 in the unitary case, we will require the following proposition. For a prime power q, let f be the unitary form on the vector space $V:={\mathbb {F}}_{\!q^2}^3$ whose Gram matrix is the $3 \times 3$ identity matrix, and let $\mathrm {SU}_3(q)$ be the associated special unitary group. Then the standard basis for (V, f) is orthonormal, and a matrix $A \in \mathrm {SL}_3(q^2)$ lies in $\mathrm {SU}_3(q)$ if and only if $A^{-1} = A^{\sigma {\mathsf {T}}}$, where $\sigma$ is the field automorphism $\alpha \mapsto \alpha ^q$ of ${\mathbb {F}}_{\!q^2}$. For a subspace U of V, we will write $\mathrm {SU}_3(q)_U$ to denote the stabiliser of U in $\mathrm {SU}_3(q)$.

Proposition 2.1

Let q be a prime power greater than 2, and let X and Y be one-dimensional subspaces of the unitary space (V, f), with X non-degenerate. Then $\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y$ contains a non-scalar matrix.

Proof

We may assume without loss of generality that X contains the vector (1, 0, 0). Let (a, b, c) be a non-zero vector of Y. In addition, let $\omega$ be a primitive element of ${\mathbb {F}}_{\!q^2}$, and let $\lambda :=\omega ^{q-1}$. Then $|\lambda | = q+1 > 3$. If at least one of a, b, and c is equal to 0, then $\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y$ contains a non-scalar diagonal matrix with two diagonal entries equal to $\lambda$ and one equal to $\lambda ^{-2}$ (not necessarily in that order).

Suppose now that a, b, and c are all non-zero, and let $\mu :=b^{-1}c$. We may assume that $a = 1$. The trace map ${\alpha \mapsto \alpha + \alpha ^q}$ from ${\mathbb {F}}_{\!q^2}$ to ${\mathbb {F}}_{\!q}$ is ${\mathbb {F}}_{\!q}$-linear, and hence has a non-trivial kernel. In particular, there exists $\beta \in {\mathbb {F}}_{\!q^2}$ such that $\beta \ne 1$ and $\beta + \beta ^q = 2$. It follows from simple calculations that if $\mu ^{q+1} = -1$, then $\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y$ contains

$$\begin{aligned} \left( \begin{matrix} 1&{}0&{}0\\ 0&{}\beta &{}\mu (1-\beta ^q)\\ 0&{}\mu ^{-1}(1-\beta )&{}\beta ^q \end{matrix} \right) . \end{aligned}$$

If instead $\mu ^{q+1} \ne -1$, then we can define $\gamma :=\lambda ^{-2}(\lambda ^3+\mu ^{q+1})(1+\mu ^{q+1})^{-1}$. In this case, $\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y$ contains

$$\begin{aligned} \left( \begin{matrix} \lambda &{}0&{}0\\ 0&{}\gamma &{}\mu (\lambda -(\gamma \lambda )^q)\\ 0&{}\mu ^{-1}(\lambda -\gamma )&{}(\lambda \gamma )^q \end{matrix} \right) . \end{aligned}$$

Note that $\lambda \ne \gamma$ since $|\lambda | > 3$. $\square$

Proof of Theorem 1.1

Let $S_1$ and $S_2$ be proper non-trivial subgroups of G, and let $M_1$ and $M_2$ be maximal subgroups of G that contain $S_1$ and $S_2$, respectively. Since $d(M_1,M_2) \leqslant d(S_1,M_2) \leqslant d(S_1,S_2)$, we may assume that $S_1$ and $S_2$ are not maximal in G. We may also assume that $M_1 \ne M_2$, as otherwise $S_1 \sim M_1 \sim S_2$ and $d(S_1,S_2) \leqslant 2$.

Suppose first that $|M_1|$ and $|M_2|$ are even. Then, as observed in the proof of [7, Proposition 3.1], there exist involutions $x \in M_1$ and $y \in M_2$, with $\langle x,y \rangle$ equal to a (proper) dihedral subgroup D of G (with $|D| = 2$ allowed). Hence $S_1 \sim M_1 \sim D \sim M_2 \sim S_2$, and so $d(S_1,S_2) \leqslant 4$. In particular, if every maximal subgroup of G has even order, then $\mathrm {diam}(\Delta _G) \leqslant 4$, as noted in the proof of [12, Lemma 2.3].

It remains to consider the case where G contains a maximal subgroup of odd order. Liebeck and Saxl [11, Theorem 2] present a list containing all possibilities for G and its maximal subgroups of odd order. By the previous paragraph, we may assume that the maximal subgroup $M_1$ has odd order. However, $|M_2|$ may be even. In what follows, information about the sporadic simple groups is taken from the Atlas [4], except where specified otherwise.

(i) $G = A_p$, with p prime, $p \equiv 3 \pmod 4$, and $p \notin \{7,11,23\}$. By [5, Theorem 2] (see also [14, Assertion I]), the intersection graph of any simple alternating group has diameter at most 4.

(ii) $G = \mathrm {L}_2(q)$, with q a prime power and $q \equiv 3 \pmod 4$. The group G acts transitively on the set $\Omega$ of one-dimensional subspaces of the vector space ${\mathbb {F}}_{\!q}^2$. Additionally, $M_1 = G_U$ for some $U \in \Omega$, and $G_U \cap G_W \ne 1$ for each $W \in \Omega$. If $|M_2|$ is odd, then $M_2 = G_W$ for some W, and it follows that $M_1 \sim M_2$ and $d(S_1,S_2) \leqslant 3$. We may therefore assume that $M_2$ contains an involution g. Then g fixes no subspace in $\Omega$, and so $g \in G_{\{U,X\}}$ for some $X \in \Omega \setminus \{U\}$. Since the non-trivial subgroup $G_U \cap G_X$ lies in both $M_1 = G_U$ and $G_{\{U,X\}}$, we deduce that $S_1 \sim M_1 \sim G_{\{U,X\}} \sim M_2 \sim S_2$. Thus $d(S_1,S_2) \leqslant 4$.

(iii) $G = \mathrm {L}_n(q)$, with n an odd prime, q a prime power, and $G \not \cong \mathrm {L}_3(4)$. Similarly to the previous case, the group G and its overgroup $R:=\mathrm {PGL}_n(q)$ act transitively on the set $\Omega$ of one-dimensional subspaces of the vector space ${\mathbb {F}}_{\!q}^n$. Here, $M_1 = {G \cap N_R(K)}$, where K is a Singer subgroup of R, i.e., a cyclic subgroup of order $(q^n-1)/(q-1)$ (see [8, §1–2]).

Now, $M_1$ contains a non-identity element m that fixes a subspace $X \in \Omega$ [8, p. 497]. Observe that ${m^k \in M_1}$ for each $k \in K$. The action of K on $\Omega$ is transitive, and hence each subspace in $\Omega$ is fixed by some non-identity element of $M_1$. Therefore, if a non-identity element of $S_2$ fixes a subspace $U \in \Omega$, then $S_1 \sim M_1 \sim G_U \sim S_2$ and $d(S_1,S_2) \leqslant 3$. Otherwise, since n is prime, there exists $g \in G$ such that $S_2 \cap M_1^g \ne 1$. Thus ${S_1 \sim M_1 \sim G_X \sim M_1^g \sim S_2}$ and $d(S_1,S_2) \leqslant 4$.

(iv) $G = \mathrm {U}_n(q)$, with n an odd prime, q a prime power, and $G \not \cong \mathrm {U}_3(3)$, $\mathrm {U}_3(5)$, or $\mathrm {U}_5(2)$. Here, G acts intransitively on the set of one-dimensional subspaces of the vector space ${\mathbb {F}}_{\!q^2}^n$. Let $(q+1,n)$ denote the greatest common divisor of $q+1$ and n. The maximal subgroup $M_1$ is equal to $N_G(T)$, where T is a Singer subgroup of G, i.e., a cyclic subgroup of order $\frac{q^n+1}{(q+1)(q+1,n)}$ (see [8, §5]). In fact, each maximal subgroup of G of odd order is conjugate to $M_1$. Similarly to the linear case, $M_1$ contains a non-identity element that fixes a one-dimensional subspace X of ${\mathbb {F}}_{\!q^2}^n$ [8, p. 512].

Let $L:=G_X$. Then $M_1 \sim L$, and we can calculate |L| using [3, Table 2.3]. In particular, |L| is even. Hence if $|M_2|$ is even, then G contains a dihedral subgroup D such that ${S_1 \sim M_1 \sim L \sim D \sim M_2 \sim S_2}$, and $d(S_1,S_2) \leqslant 5$. If $|M_2|$ is odd, then there exists an element $g \in G$ such that $M_2 = M_1^g$. Thus $L^g \sim M_2$. If $n = 3$ and X is non-degenerate, then it follows from Proposition 2.1 that $L \sim L^g$. Therefore, ${S_1 \sim M_1 \sim L \sim L^g \sim M_2 \sim S_2}$ and $d(S_1,S_2) \leqslant 5$. In the remaining cases, we will show that $|L|^2/|G| > 1$, and hence $|L|\,|L^g| > |G|$. It will follow that $L \cap L^g \ne 1$, again yielding $d(S_1,S_2) \leqslant 5$.

Observe that $|L|^2/|G| > 1$ if and only if $\log |G|/\log |G:L| > 2$. By [6, Proposition 3.2], if $n \geqslant 7$, then $\log |G|/\log |G:L| > 2$, as required. If instead $n = 3$, then we may assume that X is totally singular. Here, $q > 2$, and hence

$$\begin{aligned} |L|^2/|G| = \frac{q^3(q^2-1)}{(q^3+1)(q+1,3)} \geqslant \frac{q^3(q-1)}{(q^3+1)} > 1. \end{aligned}$$

Suppose finally that $n = 5$. If X is totally singular, then $|L|^2/|G|$ is equal to

$$\begin{aligned} \frac{q^{10}(q^2-1)^3(q^3+1)}{(q^4-1)(q^5+1)(q+1,5)}> \frac{q^{10}}{(q^4-1)(q^5+1)} = \frac{q^{10}}{q^9-q^5+q^4-1} > 1. \end{aligned}$$

If instead X is non-degenerate, then

$$\begin{aligned} |L|^2/|G| = \frac{q^2(q+1)\prod _{i=1}^{4}(q^i-(-1)^i)}{(q^5+1)(q+1,5)}> \frac{q^2(q^{4}-1)}{q^5+1} = \frac{q^6-q^2}{q^5+1} > 1. \end{aligned}$$

(v) $G = \mathrm {M}_{23}$. In this case, $M_1$ has shape $23 :11$. We argue as in the proof of [14, Assertion I]. There exists a maximal subgroup L of G isomorphic to $\mathrm {M}_{22}$, and $|M_1|\,|L|$ and $|M_2|\,|L|$ are greater than |G| (for any choice of $M_2$). It follows that $S_1 \sim M_1 \sim L \sim M_2 \sim S_2$, and so $d(S_1,S_2) \leqslant 4$.

(vi) $G = \mathrm {Th}$. Here, $M_1$ has shape $31 :15$. If the proper non-trivial subgroup $S_1$ of $M_1$ has order 31, then $S_1$ lies in a maximal subgroup of shape $2^5\cdot \mathrm {L}_5(2)$. Otherwise, $|C_G(S_1)|$ is even. Therefore, in each case, $S_1$ lies in a maximal subgroup of even order. The same is true for $S_2$, and thus $d(S_1,S_2) \leqslant 4$.

(vii) $G = {\mathbb {B}}$. In this case, $M_1$ has shape $47 :23$. Additionally, G has a maximal subgroup $K \cong \mathrm {Fi}_{23}$, which has even order, and $M_1 \sim K$. Hence if $|M_2|$ is even, then $S_1 \sim M_1 \sim K \sim D \sim M_2 \sim S_2$ for some dihedral subgroup D of G, yielding $d(S_1,S_2) \leqslant 5$. Otherwise, there exists an element $g \in G$ such that $M_2 = M_1^g$, and hence $K^g \sim M_2$. As $|K|^2/|G| > 1$, we conclude that ${S_1 \sim M_1 \sim K \sim K^g \sim M_2 \sim S_2}$ and $d(S_1,S_2) \leqslant 5$. Thus $\mathrm {diam}(\Delta _G) \leqslant 5$.

We now show that $\mathrm {diam}(\Delta _G)$ is equal to 5. Let H be a subgroup of $M_1$ of order 23. Then H is a Sylow subgroup of G. It follows from [15, p. 67] that each maximal subgroup of G that contains H is conjugate either to $M_1$, to K, or to a subgroup L of shape $2^{1+22}\cdot \mathrm {Co}_2$. We may assume that $H \leqslant M_1 \cap K \cap L$. Additionally, $N_G(H)$ has shape $(23 :11) \times 2$ and $N_L(H) = N_G(H)$, while ${|N_G(H):N_{M_1}(H)|} = 22$. Since the 22 non-identity elements of H fall into two K-conjugacy classes and $C_K(H) = H$, we conclude that $N_K(H)$ has shape $23 :11$, and so ${|N_G(H):N_K(H)|} = 2$.

Consider the pairs $(H',M')$, where $H'$ is a G-conjugate of H, $M'$ is a G-conjugate of $M_1$, and $H' \leqslant M'$. As any two G-conjugates of H appear in an equal number of such pairs, we deduce that H lies in exactly ${|N_G(H):N_{M_1}(H)|} = 22$ G-conjugates of $M_1$. Similarly, H lies in two G-conjugates of K and one G-conjugate of L.

As $M_1$ has shape $47 :23$, it contains a subgroup S of order 47. In fact, $M_1$ is the unique maximal subgroup of G that contains S. Hence if J is a maximal subgroup of G satisfying $J \ne M_1$ and $J \cap M_1 \ne 1$, then J contains a G-conjugate of H. Let ${\mathcal {U}}$ be the set of G-conjugates of H that lie in at least one such maximal subgroup J, or in $M_1$. There are 47 subgroups of order 23 in $M_1$, each of which lies in two G-conjugates of K, and there are $|K:N_K(H)|$ subgroups of order 23 in K. Therefore, there are fewer than $47 \cdot 2|K:N_K(H)|$ subgroups in ${\mathcal {U}}$ that lie in at least one G-conjugate of K. By considering the G-conjugates of $M_1$ and L similarly, we conclude that

$$\begin{aligned} |{\mathcal {U}}|< 47({2|K:N_K(H)|}+ 22\cdot 47 + |L:N_L(H)|) < |G:M_1|/22. \end{aligned}$$

Hence there exists $g \in G$ such that no subgroup of $M_1^g$ lies in ${\mathcal {U}}$. This means that $M_1$ and $M_1^g$ are not adjacent in $\Delta _G$ and have no common neighbours, and so $d(M_1,M_1^g) > 2$. As $M_1$ and $M_1^g$ are the unique neighbours of S and $S^g$, respectively, it follows that $d(S,S^g) > 4$. Therefore, $\mathrm {diam}(\Delta _G) = 5$.

(viii) $G = {\mathbb {M}}$. Liebeck and Saxl list two possible maximal subgroups of odd order (up to conjugacy), of shape $59 :29$ and $71 :35$, respectively. However, these subgroups are not, in fact, maximal: the former lies in the maximal subgroup $\mathrm {L}_2(59)$ constructed in [9], and the latter lies in the maximal subgroup $\mathrm {L}_2(71)$ constructed in [10]. Hence G has no maximal subgroup of odd order, and so $\mathrm {diam}(\Delta _G) \leqslant 4$. $\square$