On the Number of Cyclic Subgroups of a Finite Group

Abstract

Let G be a finite group and let c(G) be the number of cyclic subgroups of G. We study the function $\alpha (G) = c(G)/|G|$. We explore its basic properties and we point out a connection with the probability of commutation. For many families $\mathscr {F}$ of groups we characterize the groups $G \in \mathscr {F}$ for which $\alpha (G)$ is maximal and we classify the groups G for which $\alpha (G) > 3/4$. We also study the number of cyclic subgroups of a direct power of a given group deducing an asymptotic result and we characterize the equality $\alpha (G) = \alpha (G/N)$ when G / N is a symmetric group.

1 Introduction

In this paper all the groups we consider are finite. Let c(G) be the number of cyclic subgroups of a group G and $\alpha (G) := c(G)/|G|$. It is clear that $0 < \alpha (G) \le 1$. Observe that every cyclic subgroup $\langle x \rangle$ of G has $\varphi (o(x))$ generators, where $\varphi$ is Euler’s totient function and o(x) denotes the order of the element x, hence

$$\begin{aligned} c(G) = \sum _{x \in G} \frac{1}{\varphi (o(x))}. \end{aligned}$$

On a computational level this formula is probably best for computing c(G) for arbitrary G, and it is what we employed to work out small groups using (The GAP Group 2018). For d a divisor of $|G|=n$ let $B_G(d)$ be the number of elements $x \in G$ such that $x^d=1$. Denote by $\mu$ the standard Möbius function. In Garonzi and Patassini (2017) another formula is given (from Lemma 7 (2) choosing A(d) the number of elements of G of order d, $B(d)=B_G(d)$ and $(r,s)=(1,0)$), which is an easy application of the Möbius inversion formula, the following:

$$\begin{aligned} c(G) = \sum _{d|n} \left( \sum _{i|n/d} \frac{\mu (i)}{\varphi (di)} \right) B_G(d). \end{aligned}$$

(1)

Using this formula in (Garonzi and Patassini 2017, Corollary 13) it was shown that ciclicity can be detected by the number of cyclic subgroups, more precisely that if $|G|=n$ then $c(G) \ge c(C_n)$ with equality if and only if $G \cong C_n$.

There is a connection between $\alpha (G)$ and the so-called “commuting probability” of G, denoted by ${{\mathrm{cp}}}(G)$, that is the probability that two random elements of G commute (studied extensively in Guralnick and Robinson (2006), which we crucially employ in our study). More specifically we prove that if $\alpha (G) \ge 1/2$ then ${{\mathrm{cp}}}(G) \ge (2 \alpha (G)-1)^2$. This implies that a group with many cyclic subgroups has big solvable radical and, if it is already solvable, it has big Fitting subgroup (see Sect. 6 for the details).

It is not hard to show that $\alpha (G) \le \alpha (G/N)$ whenever N is a normal subgroup of G, and if equality holds then N must be an elementary abelian 2-group. It is an interesting question to ask what can we say about G if $\alpha (G)=\alpha (G/N)$ given some information on G / N. In this paper we characterize equality $\alpha (G) = \alpha (G/N)$ when G / N is a symmetric group (Theorem 1).

Given a family $\mathscr {F}$ of groups define

$$\begin{aligned} \alpha _{\mathscr {F}} := \max \{\alpha (G)\ :\ G \in \mathscr {F}\},\qquad m\mathscr {F} := \{G \in \mathscr {F}\ :\ \alpha (G) = \alpha _{\mathscr {F}}\}. \end{aligned}$$

We are interested in computing $\alpha _{\mathscr {F}}$ and $m\mathscr {F}$ for various families $\mathscr {F}$. In this paper we prove the following results.

• 1. If $\mathscr {F}$ is the family of all finite groups then $\alpha _{\mathscr {F}}=1$ and $m\mathscr {F}$ is the family of elementary abelian 2-groups (by 2.2).

• 2. If $\mathscr {F}$ is the family of non-abelian groups then

  $$\begin{aligned} \alpha _{\mathscr {F}} = 7/8 = \alpha (D_8) \end{aligned}$$

  and $m \mathscr {F}$ is the family of groups of the form ${C_2}^n \times D_8$ for some $n \ge 0$ (by Corollary 2).

• 3. If $\mathscr {F}$ is the family of non-nilpotent groups then

  $$\begin{aligned} \alpha _{\mathscr {F}} = 5/6 = \alpha (S_3) \end{aligned}$$

  and $m \mathscr {F}$ is the family of groups of the form ${C_2}^n \times S_3$ for some $n \ge 0$ (by Corollary 2).

• 4. If $\mathscr {F}$ is the family of non-solvable groups then

  $$\begin{aligned} \alpha _{\mathscr {F}} = 67/120 = \alpha (S_5) \end{aligned}$$

  and $m \mathscr {F}$ is the family of groups of the form ${C_2}^n \times S_5$ for some $n \ge 0$ (Theorem 4).

• 5. If $\mathscr {F}$ is the family of non-supersolvable groups then

  $$\begin{aligned} \alpha _{\mathscr {F}} = 17/24 = \alpha (S_4) \end{aligned}$$

  and $m \mathscr {F}$ is the family of groups of the form ${C_2}^n \times S_4$ for some $n \ge 0$ (Theorem 6).

• 6. If p is an odd prime and $\mathscr {F}$ is the family of non-trivial groups of order divisible only by primes at least p then

  $$\begin{aligned} \alpha _{\mathscr {F}} = 2/p = \alpha (C_p) \end{aligned}$$

  and $m \mathscr {F} = \{C_p\}$ (Proposition 1).

• 7. If p is an odd prime and $\mathscr {F}$ is the family of groups G with $C_p$ as an epimorphic image then

  $$\begin{aligned} \alpha _{\mathscr {F}}=2/p=\alpha (C_p) \end{aligned}$$

  and $m \mathscr {F}$ is the family of groups which are the direct product of an elementary abelian 2-group and a Frobenius group with 2-elementary abelian Frobenius kernel and Frobenius complements of order p (Proposition 2).

We also classify the groups G with $\alpha (G) > 3/4$ (Theorem 5), proving in particular that 3 / 4 is the largest non-trivial accumulation point of the set of numbers of the form $\alpha (G)$. An easy consequence of this (Corollary 3) is that if G is not an elementary abelian 2-group and it has $|G|-n$ cyclic subgroups then $|G| \le 8n$. This extends and generalizes the results in Tarnauceanu (2015), as we show right after the corollary. We also give a formula for $\alpha (G)$ when G is a nilpotent group (Theorem 2) and we study $\alpha$ of a direct power (Theorem 3) proving that $G^n$ has roughly $|G^n|/\varphi (\exp (G))$ cyclic subgroups.

2 Basic Properties of $\alpha$

In this section we prove some basic properties of the function $\alpha$.

2.1. If A and B are finite groups of coprime orders then $c(A \times B) = c(A) c(B)$ and hence $\alpha (A \times B) = \alpha (A) \alpha (B)$. The proof of this is straightforward.

2.2. Let I(G) denote the number of elements $g \in G$ such that $g^2=1$. Then

$$\begin{aligned} \alpha (G) \le \frac{1}{2}+\frac{I(G)}{2|G|}, \qquad \qquad \frac{I(G)}{|G|} \ge 2 \alpha (G)-1. \end{aligned}$$

In particular $\alpha (G)=1$ if and only if G is an elementary abelian 2-group.

Proof

If $g \in G$ then $g^2=1$ if and only if $\varphi (o(g))=1$, so

$$\begin{aligned} c(G) = \sum _{x \in G} \frac{1}{\varphi (o(x))} \le I(G)+\frac{1}{2}(|G|-I(G)) = \frac{1}{2}(I(G)+|G|). \end{aligned}$$

This implies the result. $\square$

2.3. If $N \unlhd G$ then $\alpha (G) \le \alpha (G/N)$. Moreover $\alpha (G) = \alpha (G/N)$ if and only if $\varphi (o(g)) = \varphi (o(gN))$ for every $g \in G$, where o(gN) denotes the order of the element gN in the group G / N.

Proof

If a divides b then $\varphi (a) \le \varphi (b)$, therefore

$$\begin{aligned} c(G/N)&= \sum _{gN \in G/N} \frac{1}{\varphi (o(gN))} = \sum _{g \in G} \frac{1}{|N| \varphi (o(gN))} \ge \sum _{g \in G} \frac{1}{|N| \varphi (o(g))} = \frac{c(G)}{|N|}. \end{aligned}$$

This implies the result. $\square$

2.4. If $\alpha (G) = \alpha (G/N)$ then N is an elementary abelian 2-group.

Proof

If $n \in N$ then applying 2.3 we have $\varphi (o(n)) = \varphi (o(nN)) = \varphi (o(N)) = \varphi (1)=1$ so $n^2=1$. $\square$

2.5. If G is any finite group then $\alpha (G) = \alpha (G \times {C_2}^n)$ for all $n \ge 0$.

Proof

Choosing $N = \{1\} \times {C_2}^n$ gives $\varphi (o(x)) = \varphi (o(xN))$ for all $x \in G \times {C_2}^n$. The result follows from 2.3. $\square$

2.6. If $\alpha (G) = \alpha (G/N)$ and $L \unlhd G$, $L \subseteq N$ then $\alpha (G) = \alpha (G/L)$.

Proof

Since G / N is a quotient of G / L we have $\alpha (G/N) = \alpha (G) \le \alpha (G/L) \le \alpha (G/N)$ by 2.3 and the result follows. $\square$

2.7. If $\alpha (G) = \alpha (G/N)$ and $K \le G$ then $\alpha (K) = \alpha (K/K \cap N)$.

Proof

Let $R := K \cap N$. By 2.3, is enough to show that if $x \in K$ then $o(xR)=o(xN)$ (because then $\varphi (o(xR))=\varphi (o(xN))=\varphi (o(x))$). Let $a=o(xR)$ and $b=o(xN)$. Since $R \subseteq N$ we have $x^a \in N$ so $b \le a$. On the other hand $x^b \in K \cap N = R$ so $a \le b$. Therefore $a=b$. $\square$

2.8. Suppose $\alpha (G) = \alpha (G/N)$. If $a \in G$ has order 2 modulo N then a centralizes N, in particular if G / N can be generated by elements of order 2 then $N \subseteq Z(G)$.

Proof

We have $\varphi (o(a)) = \varphi (o(aN)) = \varphi (2) = 1$ by 2.3, and a has order 2 modulo N, so $o(a)=2$. If $n \in N$ then $\varphi (o(an)) = \varphi (o(anN)) = \varphi (o(aN)) = 1$ so $(an)^2=1$. This together with $a^2=n^2=1$ (by 2.4) implies $an=na$. Recalling that N is abelian (by 2.4) we deduce that if G / N can be generated by elements of order 2 then $N \subseteq Z(G)$. $\square$

3 A Characterization

Observe that $C_3$ is a quotient of $A_4$ and $\alpha (C_3) = \alpha (A_4) = 2/3$, so it is not always the case that $\alpha (G) = \alpha (G/N)$ implies $G \cong N \times G/N$. We can characterize the groups such that $\alpha (G) = \alpha (G/N)$ when G / N is a symmetric group, for the following two reasons: the symmetric groups can be generated by elements of order 2 and their double covers are known.

Theorem 1

Let G be a group and N a normal subgroup of G such that G / N is isomorphic to a symmetric group. If $\alpha (G) = \alpha (G/N)$ then N is an elementary abelian 2-group and it admits a normal complement in G, so that $G \cong N \times G/N$.

Proof

We prove the result by induction on the order of G. By 2.4, N is an elementary abelian 2-group. Since $G/N \cong S_m$ can be generated by elements of order 2, 2.8 implies that N is central in G. If $m=2$ then the result follows from 2.2, so suppose $m \ge 3$. Let $R \cong {C_2}^l$ be a minimal normal subgroup of G contained in N. By 2.6 we have $\alpha (G/R)=\alpha (G)=\alpha (S_m)$ so by induction, since G / N is a quotient of G / R, we have $G/R = {C_2}^l \times S_m$ for some $l \ge 0$. Let $K \unlhd G$ be the (normal) subgroup of G such that $K/R = \{1\} \times S_m$. Observe that $K \cap N$ contains R, so $K \cap N/R$ is a normal 2-subgroup of $K/R \cong S_m$. If $m \ne 4$ this implies that $K \cap N = R$ because $S_m$ in this case does not admit non-trivial normal 2-subgroups (being $m \ge 3$). If $m=4$ and $K \cap N \ne R$ then $K \cap N/R$ is the Klein group, and $K/K \cap N \cong S_3$. However in this case 2.3 and 2.7 imply that

$$\begin{aligned} 17/24&= \alpha (S_4) = \alpha (K/R) \ge \alpha (K) = \alpha (K/K \cap N) = \alpha (S_3) = 5/6, \end{aligned}$$

a contradiction. We deduce that $K \cap N = R$.

If $N \ne R$ then $|K| < |G|$ and $\alpha (K) = \alpha (K/R) = \alpha (S_m)$ by 2.7 (being $K \cap N = R$). By induction we deduce that $K \cong R \times S_m$. Set $M := \{1\} \times S_m \le K$. Since

$$\begin{aligned} S_m \cong G/N \ge KN/N \cong K/K \cap N = K/R \cong S_m \end{aligned}$$

we obtain $G=KN=MRN=MN$ so being N central in G and $N \cap M = N \cap K \cap M = R \cap M = \{1\}$ we deduce $G = N \times M \cong N \times G/N$. Assume now $N=R$, so N is a minimal normal subgroup of G. Since N is central, $|N|=2$ and actually $N = \langle z \rangle = Z(G)$ is the center of G (being $G/N \cong S_m$ with $m \ge 3$). Suppose by contradiction that G is not a direct product $C_2 \times S_m$. We claim that N is contained in the derived subgroup of G. Indeed $G'$ is contained in the subgroup T of G such that $T \supseteq N$, $T/N \cong A_m$ (being $|G/T|=2$), so if $G'$ does not contain N then $G'N/N$ is a nontrivial normal subgroup of $G/N \cong S_m$ containing the derived subgroup of $S_m$ (that is, $A_m$) hence $G'N = T$ therefore letting $\varepsilon \in G$ represent a fixed element of order 2 of $G/N \cong S_m$ not belonging to $A_m$, $\varphi (o(\varepsilon )) = \varphi (2) = 1$ (by 2.3) hence $o(\varepsilon )=2$ implying that $G' \langle \varepsilon \rangle \cap N = \{1\}$ (otherwise $G' \langle \varepsilon \rangle \supseteq N$ implying that $G' \langle \varepsilon \rangle = G$ so $|G:G'|=2$ hence $G \cong N \times G'$, a contradiction) therefore $G' \langle \varepsilon \rangle \cong G' \langle \varepsilon \rangle N/N = G/N \cong S_m$; being N the center of G we deduce $G \cong C_2 \times S_m$, a contradiction. This implies that $N \subseteq G'$ so G is a double cover of $S_m$ (that is, a stem extension of $S_m$ where the base normal subgroup has order 2), and looking at the known presentations of the double covers of the symmetric group (classified by Schur, see for example Schur (2001)) we see that z is a square in G, that is there exists $x \in G$ with $x^2=z$, so that x has order 4 and xN has order 2, contradicting $\varphi (o(x)) = \varphi (o(xN))$ (which is true by 2.3). $\square$

4 Nilpotent Groups

Let G be a finite group. For $\ell$ a divisor of |G| let $B_G(\ell )$ be the number of elements $g \in G$ with the property that $g^{\ell }=1$ and let $r_G(\ell )$ be the number of elements of G of order $\ell$. It is worth mentioning the famous result by Frobenius that if G is any group and $\ell$ divides |G| then $\ell$ divides $B_G(\ell )$. The idea of the following result, which is a reformulation of formula (1) in the nilpotent case, is to give a formula for c(G) when G is a nilpotent group in terms of the numbers $B_G(d)$, that in general are reasonably easy to deal with (consider for example the case in which G is abelian).

Theorem 2

If G is a nilpotent group of order n then $c(G) = \sum _{d|n} B_G(d)/d$.

Proof

Assume first that G is a p-group, $|G|=p^n$. Since $r_G(p^j)=B_G(p^j)-B_G(p^{j-1})$ whenever $j \ge 1$ we see that

$$\begin{aligned} c(G)&= \sum _{x \in G} \frac{1}{\varphi (o(x))} = \sum _{j=0}^n \frac{r_G(p^j)}{\varphi (p^j)} = 1+\sum _{j=1}^{n} \frac{B_G(p^j)-B_G(p^{j-1})}{\varphi (p^j)} \\&= 1-\frac{B_G(1)}{\varphi (p)}{+}\sum _{j=1}^{n-1} \left( \frac{1}{\varphi (p^j)}-\frac{1}{\varphi (p^{j+1})} \right) B_G(p^j)+\frac{B_G(p^n)}{\varphi (p^n)} {=} \sum _{j=0}^n \frac{B_G(p^j)}{p^j}. \end{aligned}$$

Now consider the general case, and write the order of G as $n=|G|={p_1}^{n_1} \cdots {p_t}^{n_t}$ with the $p_i$’s pairwise distinct primes, G is a direct product $\prod _{i=1}^t G_{p_i}$ where $G_{p_i}$ is the unique Sylow $p_i$-subgroup of G. Using 2.1 we obtain

$$\begin{aligned} c(G) = \prod _{i=1}^t c(G_{p_i}) = \prod _{i=1}^t \left( \sum _{j=0}^n \frac{B_G(p_i^j)}{p_i^j} \right) = \sum _{d|n} B_G(d)/d. \end{aligned}$$

The last equality follows from the fact that since G is nilpotent $B_G(ab)$ equals $B_G(a)B_G(b)$ if a and b are coprime divisors of n. $\square$

5 An Asymptotic Result

We want to study $\alpha (G^n)$ where $G^n = G \times G \times \cdots \times G$ (n times) in terms of the functions $B_G$ and $r_G$ defined in the previous section. Recall that the exponent of a group G, denoted $\exp (G)$, is the least common multiple of the orders of the elements of G. It is clear that $\exp (G^n)=\exp (G)$. The following result shows that $G^n$ has roughly $|G^n|/\varphi (\exp (G))$ cyclic subgroups.

Theorem 3

Let G be a finite group. Then $\lim _{n \rightarrow \infty } \alpha (G^n) = 1/\varphi (\exp (G))$.

Proof

Observe that $r_{G^n}(\ell ) \ne 0$ only if $\ell$ divides |G|, therefore

$$\begin{aligned} \alpha (G^n) = \frac{1}{|G^n|} \sum _{x \in G^n} \frac{1}{\varphi (o(x))} = \sum _{\ell | |G|^n} \frac{r_{G^n}(\ell )}{\varphi (\ell )|G|^n} = \sum _{\ell | |G|} \frac{r_{G^n}(\ell )}{\varphi (\ell )|G|^n} \end{aligned}$$

so what we need to compute is the limit $L_{\ell }$ of $r_{G^n}(\ell )/|G|^n$ when $n \rightarrow \infty$, for $\ell$ a divisor of |G|. Clearly $r_{G^n}(\ell ) \le B_{G^n}(\ell ) = B_G(\ell )^n$ hence $r_{G^n}(\ell )/|G|^n \le (B_G(\ell )/|G|)^n$ so if $B_G(\ell ) < |G|$ then $L_{\ell }=0$. Now assume $B_G(\ell )=|G|$, in other words $\exp (G)$ divides $\ell$. If $\exp (G) < \ell$ then $r_{G^n}(\ell )=0$ so $L_{\ell }=0$. Now assume $\exp (G) = \ell$. Let p vary in the set of prime divisors of |G|, and for every such p define $a_p := B_G(\exp (G)/p)$. Clearly $G^n$ has at least $|G|^n-\sum _p {a_p}^n$ elements of order $\exp (G)$. Observe that $a_p < |G|$ by definition of $\exp (G)$, so that $a_p/|G| < 1$, hence $(a_p/|G|)^n$ tends to 0 as $n \rightarrow \infty$, implying $L_{\exp (G)}=1$. The result follows. $\square$

6 A Connection with the Probability of Commutation

The probability that two elements in a group G commute is denoted by ${{\mathrm{cp}}}(G)$ (“commuting probability” of G) and is defined by $|S|/|G \times G|$ where S is the set of pairs $(x,y) \in G \times G$ such that $xy=yx$. It is easy to show that ${{\mathrm{cp}}}(G)=k(G)/|G|$ where k(G) is the number of conjugacy classes of G. This invariant was studied by many authors, but we refer mostly to Guralnick and Robinson (2006).

Let I(G) be the size of the set $\{x \in G\ :\ x^2=1\}$. The following lemma is easily deducible from Theorem 2J of Brauer and Fowler (1955). It can also be proved character-theoretically using the Frobenius-Schur indicator.

Lemma 1

$I(G)^2 \le k(G)|G|$, in other words ${{\mathrm{cp}}}(G) \ge (I(G)/|G|)^2$.

This together with 2.2 implies the following inequality.

Lemma 2

If $\alpha (G) \ge 1/2$ then ${{\mathrm{cp}}}(G) \ge (2\alpha (G)-1)^2$.

Let us include some other results from Guralnick and Robinson (2006) that we will need in the following section.

6.1. If G is a non-solvable group and ${{\mathrm{sol}}}(G)$ is the maximal normal solvable subgroup of G then ${{\mathrm{cp}}}(G) \le |G:{{\mathrm{sol}}}(G)|^{-1/2}$. This follows from (Guralnick and Robinson 2006, Theorem 9) (which depends on the classification of the finite simple groups), and together with Lemma 2 implies that if $\alpha (G) > 1/2$ then $|G:{{\mathrm{sol}}}(G)| \le (2\alpha (G)-1)^{-4}$.

6.2. If G is a solvable group and F(G) is the Fitting subgroup of G then ${{\mathrm{cp}}}(G) \le |G:F(G)|^{-1/2}$. This follows from (Guralnick and Robinson 2006, Theorem 4), and together with Lemma 2 implies that if $\alpha (G) > 1/2$ then $|G:F(G)| \le (2\alpha (G)-1)^{-4}$.

7 Non-Solvable Groups

Theorem 4

Let G be a finite non-solvable group. Then $\alpha (G) \le \alpha (S_5)$ with equality if and only if $G \cong S_5 \times {C_2}^n$ for some integer $n \ge 0$.

Proof

Let $\alpha := \alpha (S_5) = 67/120$. We will show that if G is any finite non-solvable group such that $\alpha (G) \ge \alpha$ then $G \cong S_5 \times {C_2}^n$ for some $n \ge 0$. Assume $\alpha (G) \ge \alpha$, in particular $\alpha (G) > 1/2$. By 6.1 we deduce $|G/{{\mathrm{sol}}}(G)| \le (2 \alpha -1)^{-4} = (60/7)^4 < 5398$ thus $|G/{{\mathrm{sol}}}(G)| \le 5397$. Observe that $G/{{\mathrm{sol}}}(G)$ is non-trivial (being G non-solvable), it does not have non-trivial solvable normal subgroups and $\alpha (G/{{\mathrm{sol}}}(G)) \ge \alpha (G) \ge \alpha$. If we can show that $G/{{\mathrm{sol}}}(G) \cong S_5$ it will follow that $67/120 = \alpha \le \alpha (G) \le \alpha (G/{{\mathrm{sol}}}(G)) = \alpha (S_5) = 67/120$ therefore $\alpha (G) = \alpha (G/{{\mathrm{sol}}}(G))$ and the result follows from Theorem 1.

We are left to show that if G is a group without non-trivial solvable normal subgroups and such that $|G| \le 5397$ and $\alpha (G) \ge \alpha (S_5) = 67/120$, then $G \cong S_5$. Let N be a minimal normal subgroup of G, then $N=S^t$ with S a non-abelian simple group. If $t \ge 2$ then being $|S| \ge 60$ and $|G| \le 5397$ we deduce $G=N=A_5 \times A_5$, contradicting the minimality of N. So $t=1$. We claim that there is no other minimal normal subgroup of G. Indeed if M is a minimal normal subgroup of G distinct from N then M is non-solvable (by assumption) so $|G:MN| = |G|/|MN| \le 5397/60^2 < 2$ (the smallest order of a non-solvable group is 60) so $G=MN$ and actually $G = M \times N = A_5 \times A_5$ (M is a direct power of a non-abelian simple group, the smallest orders of non-abelian simple groups are 60, 168 and $60 \cdot 60^2,60 \cdot 168$ are both larger than 5397) which is a contradiction because $\alpha (A_5 \times A_5) = 77/225 < \alpha (S_5)$. We deduce that N is the unique minimal normal subgroup of G. Since N is non-solvable, it is non-abelian, so it is not contained in $C_G(N)$, hence $C_G(N)$ must be trivial (otherwise it would contain a minimal normal subgroup of G distinct from N) therefore G is almost-simple. Using Leemans and Vauthier (2006) and The GAP Group (2018) we computed the list of almost-simple groups G of size at most 5397 and for each of them we determined $\alpha (G)$. The results are summarized in the Table 1. We deduce that the only almost-simple group G with $|G| \le 5397$ and $\alpha (G) \ge 67/120$ is $G = S_5$. $\square$

Table 1 Almost-simple groups of order at most 5397

This together with Lemma 2 and 6.2 implies the following.

Corollary 1

If $\alpha (G) > \alpha (S_5)$ then G is solvable and the Fitting subgroup of G has index at most 5397.

8 Groups with Many Cyclic Subgroups

In this section we will study groups with $\alpha (G)$ “large”, specifically, we will classify all the finite groups G such that $\alpha (G) > 3/4$. This is a natural choice because 3 / 4 turns out to be the largest non-trivial accumulation point of the set of numbers of the form $\alpha (G)$. To do such classification the idea is to observe that if $\alpha (G) > 3/4$ then $I(G)/|G| > 1/2$ and use Wall’s classification (Wall 1970, Section 7).

Theorem 5

Let X be a group with $\alpha (X) > 3/4$. Then X is a direct product of an elementary abelian 2-group with a group G, $\alpha (X) = \alpha (G)$, G does not have $C_2$ as a direct factor, and either G is trivial (in which case $\alpha (X)=1$) or one of the following occurs.

• (1) Case I. $G \cong A \rtimes \langle \varepsilon \rangle$, where $\langle \varepsilon \rangle = C_2$ acts on A by inversion and there exists an integer $n \ge 1$ such that one of the following occurs.

  $$\begin{aligned} A={C_3}^n,\ \alpha (G)=\frac{3 \cdot 3^n+1}{4 \cdot 3^n} \quad \text{ or } \quad A={C_4}^n,\ \alpha (G)=\frac{3 \cdot 2^n+1}{4 \cdot 2^{n}}. \end{aligned}$$

• (2) Case II. $G \cong D_8 \times D_8$ and $\alpha (G) = 25/32$.

• (3) Case III. G is a quotient ${D_8}^r/N$ where $N = \{(a_1,\ldots ,a_r) \in {Z(D_8)}^r\ :\ a_1 \cdots a_r = 1\}$ and

  $$\begin{aligned} \alpha (G) = \frac{3 \cdot 2^r+1}{4 \cdot 2^{r}}. \end{aligned}$$

• 4 Case IV. G is a semidirect product $V \rtimes \langle c \rangle$ where $V={\mathbb {F}_2}^{2r}$ has a basis $\{x_1,y_1,\ldots ,x_r,y_r\}$, c has order 2, it acts trivially on each $y_i$, $x_i^c = cx_ic = [c,x_i]x_i = x_iy_i$ for $i=1,\ldots ,r$, and

  $$\begin{aligned} \alpha (G) = \frac{3 \cdot 2^r+1}{4 \cdot 2^{r}}. \end{aligned}$$

Proof

We know by 2.5 that $\alpha (X) = \alpha (G)$. Also, we may assume that G is non-trivial. Since $\alpha (G) > 3/4$, by 2.2 we have $I(G)/|G| \ge 2 \alpha (G)-1 > 1/2$ so G appears in Wall’s classification (Wall 1970, Section 7). Case II is immediate, we will treat cases I, III and IV.

Case I of Wall’s classification. G is a semidirect product $A \rtimes \langle \varepsilon \rangle$ with A an abelian group, $\langle \varepsilon \rangle \cong C_2$ and every element of $G-A$ has order 2. Observe that A does not admit $C_2$ as a direct factor. Indeed if $a \in A$ then since $a \varepsilon \not \in A$, $a \varepsilon$ has order 2 so $a^{\varepsilon } = a^{-1}$, hence $\varepsilon$ acts on A as inversion and a direct factor of order 2 in A would yield a direct factor of order 2 in G. It follows that $c(G) = c(A)+|G|/2$, so that $3/4 < \alpha (G) = \alpha (A)/2+1/2$ implying $\alpha (A) > 1/2$. If the prime p divides the order of A then $C_p$ is a quotient of A so $1/2 < \alpha (A) \le \alpha (C_p) = 2/p$ whence $p \le 3$, that is, p is either 2 or 3. Write $A = P_2 \times P_3$ where $P_2$ is an abelian 2-group and $P_3$ is an abelian 3-group. Observe that $C_9$ is not a quotient of A because otherwise $1/2 < \alpha (A) \le \alpha (C_9) = 1/3$, a contradiction. Therefore if $P_2$ is trivial then $A = {C_3}^n$ for some $n \ge 1$, an easy computation shows $\alpha (A) = \frac{3^n+1}{2 \cdot 3^n}$, and the result follows. Suppose now that $P_2$ is non-trivial. If $P_3$ is non-trivial then since $P_2$ is not elementary abelian (because A does not have $C_2$ as a direct factor) there is a quotient of A isomorphic to $C_{12}$, however $1/2 < \alpha (A) \le \alpha (C_{12}) = 1/2$ gives a contradiction. So $P_3 = \{1\}$, in other words A is an abelian 2-group and we may write $A = \prod _{i=1}^n {C_{2^{a_i}}}$. Since A does not have $C_2$ as a direct factor we deduce $a_i \ge 2$ for all i, on the other hand if one of the $a_i$’s is at least 3 then $C_8$ is a quotient of A but $1/2 < \alpha (A) \le \alpha (C_8) = 1/2$ is a contradiction. So $A \cong {C_4}^n$ hence an easy computation shows $\alpha (A) = \frac{2^n+1}{2 \cdot 2^n}$, and the result follows.

Case III of Wall’s classification. G is a direct product of $D_8$’s with the centers amalgamated. $G=G(r)$ has a presentation

$$\begin{aligned}&G(r) = \langle c,x_1,y_1,\ldots ,x_r,y_r\ :\ c^2=x_i^2=y_i^2=1,\\&\text{ all } \text{ pairs } \text{ of } \text{ generators } \text{ commute } \text{ except } [x_i,y_i]=c \rangle . \end{aligned}$$

A more practical description of the group in question is $G = {D_8}^r/N$ where $N = \{(z_1,\ldots ,z_r) \in Z^r\ :\ z_1 \cdots z_r=1\}$ where $Z=\langle z \rangle$ (cyclic of order 2) is the center of $D_8$. N is a subgroup of $Z({D_8}^r)=Z^r$ of index 2, so $|G|=2 \cdot 4^r$. An element $(a_1,\ldots ,a_r)N \in G$ squares to 1 if and only if $(a_1^2,\ldots ,a_r^2) \in N$, that is, $a_1^2 \cdots a_r^2 = 1$. Observe that every $a_i^2$ is either 1 or z, so this condition means that there are an even number of indeces i such that $a_i^2=z$. Since $D_8$ contains 6 elements that square to 1 and 2 elements that square to z, ${D_8}^r$ contains exactly

$$\begin{aligned} \beta _r = \sum _{k=0}^{[r/2]} \left( {\begin{array}{c}r\\ 2k\end{array}}\right) 2^{2k} 6^{r-2k} \end{aligned}$$

elements $(a_1,\ldots ,a_r)$ such that $a_1^2 \cdots a_r^2 = 1$. Hence G contains exactly $\beta _r/2^{r-1}$ elements that square to 1. Observe that

$$\begin{aligned} 8^r = (2+6)^r = \sum _{h=0}^r \left( {\begin{array}{c}r\\ h\end{array}}\right) 2^h 6^{r-h}, \qquad 4^r = (-2+6)^r = \sum _{h=0}^r \left( {\begin{array}{c}r\\ h\end{array}}\right) (-1)^h 2^h 6^{r-h} \end{aligned}$$

so adding them together gives exactly $2 \beta _r$. This means that $\beta _r = \frac{1}{2} (8^r+4^r)$, so G has exactly $\beta _r/2^{r-1} = 4^r+2^r$ elements that square to 1 and exactly $|G|-\beta _r/2^{r-1} = 2 \cdot 4^r-4^r-2^r = 4^r-2^r$ elements of order 4. Therefore

$$\begin{aligned} \alpha (G) = \frac{1}{2 \cdot 4^r} \left( 4^r+2^r+\frac{1}{2} (4^r-2^r) \right) = \frac{3 \cdot 2^r+1}{4 \cdot 2^{r}}. \end{aligned}$$

Case IV of Wall’s classification. $G=G(r)$ has a presentation

$$\begin{aligned}&G(r) = \langle c,x_1,y_1,\ldots ,x_r,y_r\ :\ c^2=x_i^2=y_i^2=1,\\&\text{ all } \text{ pairs } \text{ of } \text{ generators } \text{ commute } \text{ except } [c,x_i]=y_i \rangle . \end{aligned}$$

A more practical description of the group in question is $G \cong V \rtimes \langle c \rangle$ where $V={C_2}^{2r} = \langle x_1,y_1,\ldots ,x_r,y_r \rangle$, c has order 2, it acts trivially on each $y_i$ and $x_i^c = cx_ic = [c,x_i]x_i = x_iy_i$ for $i=1,\ldots ,r$. Thinking of V as a vector space over $\mathbb {F}_2$, if $v \in V$ then vc has order 2 or 4, and it has order 4 exactly when $v^c \ne v$. Observe that with respect to the given basis of V the operator c (acting from the right) has a diagonal block matrix form with $J = \left( \begin{array}{cc} 1 &{} 1 \\ 0 &{} 1 \end{array} \right)$ on each diagonal block entry. Thus there are precisely $2^r$ vectors v with $v^c=v$, they are of the form $(0,b_1,0,b_2,\ldots ,0,b_r)$. Therefore G has $2^{2r}-2^r$ elements of order 4 and $c(G) = 2^{2r}+2^r+(2^{2r}-2^r)/2$. This implies that $\alpha (G) = \frac{3 \cdot 2^r+1}{4 \cdot 2^{r}}$. $\square$

The following corollary is immediate. It implies that if G is a non-nilpotent group then $\alpha (G) \le 5/6$ with equality if and only if G is a direct product ${C_2}^n \times S_3$.

Corollary 2

Let G be a group such that $\alpha (G) \ge 5/6 = \alpha (S_3)$. Then either

• (1) $G \cong {C_2}^n \times S_3$ for some $n \ge 0$ and $\alpha (G)=5/6$, or

• (2) $G \cong {C_2}^n \times D_8$ for some $n \ge 0$ and $\alpha (G)=7/8$, or

• (3) $G \cong {C_2}^n$ for some $n \ge 0$ and $\alpha (G)=1$.

We can deduce a bound of |G| in terms of $|G|-c(G)$. The inequality $\alpha (G) \le 7/8$ can be written as $|G| \le 8(|G|-c(G))$, so we obtain the following.

Corollary 3

If G is any finite group which is not an elementary abelian 2-group then $|G| \le 8(|G|-c(G))$ with equality if and only if $G \cong D_8 \times {C_2}^n$ for some non-negative integer n.

Observe that the above results extend and generalize the results in Tarnauceanu (2015). As an example of application let us determine the groups G with $|G|-9$ cyclic subgroups. In this case we have $|G|-c(G)=9$ so $|G| \le 72$ and a GAP check yields that G is one of $C_{11}$, $D_{22}$ and $C_4 \times S_3$.

9 Special Families of Groups

Proposition 1

Let $p \ge 3$ be a prime number. Let G be a non-trivial group of order divisible only by primes at least p. Then $\alpha (G) \le 2/p$ with equality if and only if $G \cong C_p$.

In particular if G belongs to the family of groups of odd order then $\alpha (G) \le 2/3$ with equality if and only if $G \cong C_3$.

Proof

If $1 \ne x \in G$ and q is a prime divisor of the order of x then $p \le q$ so $\varphi (o(x)) \ge \varphi (q) = q-1 \ge p-1$, so since $|G| \ge p$ we have

$$\begin{aligned} \alpha (G)&= \frac{1}{|G|} \sum _{x \in G} \frac{1}{\varphi (o(x))} \le \frac{1}{|G|} \left( 1+\frac{|G|-1}{p-1} \right) \\&= \frac{p-2}{|G|(p-1)}+\frac{1}{p-1} \le \frac{p-2}{p(p-1)}+\frac{1}{p-1} = \frac{2}{p}. \end{aligned}$$

If equality holds the above inequalities are equalities and using $p \ge 3$ it is easy to deduce that $|G|=p$, that is, $G \cong C_p$. $\square$

Proposition 2

Let G be a group and p an odd prime, and suppose G has $C_p$ as epimorphic image (in other words p divides $|G/G'|$). Then $\alpha (G) \le 2/p$ with equality if and only if G is a direct product of an elementary abelian 2-group with a Frobenius group with 2-elementary abelian kernel and complements of order p.

Observe that a bound of 2 / p when $p=2$ would be trivial. This is why we are only considering the odd case.

Proof

Since $C_p$ is a quotient of G we have $\alpha (G) \le \alpha (C_p) = 2/p$. Now assume equality holds. Then $\alpha (G) = \alpha (C_p)$ and there exists $N \unlhd G$ with $G/N \cong C_p$, so N is an elementary abelian 2-group by 2.4, say $N \cong {C_2}^m$. If G has elements of order 2p then it has $C_2$ as a direct factor (for example by Maschke’s theorem), so now assume G does not have elements of order 2p. A subgroup of G of order p acts fixed point freely on N so G is a Frobenius group with Frobenius kernel equal to N and Frobenius complement of order p. Now assume G is a Frobenius group with 2-elementary abelian kernel of size $2^m$ and Frobenius complement of order p. The element orders of G are 1, 2, and p, and G has precisely $2^m-1$ elements of order 2 and $2^m(p-1)$ elements of order p. We have

$$\begin{aligned} \alpha (G) = \frac{1}{|G|} \sum _{x \in G} \frac{1}{\varphi (o(x))} = \frac{1}{2^m \cdot p} \left( 2^m+\frac{2^m(p-1)}{p-1} \right) = 2/p. \end{aligned}$$

This concludes the proof. $\square$

10 Non-Supersolvable Groups

Let G be a solvable group and let $F_i$ be normal subgroups of G defined as follows:

$$\begin{aligned} F_0=\{1\}, \qquad F_{i+1}/F_i := F(G/F_i) \quad \forall i \ge 0, \end{aligned}$$

where $F(G/F_i)$ denotes the Fitting subgroup of $G/F_i$. In particular $F_1$ is the Fitting subgroup of G. Since G is solvable, there exists a minimal h such that $F_h=G$, such h is called the “Fitting height” of G. Observe that

$$\begin{aligned} F(F_l/F_{i-1}) = F_i/F_{i-1} \qquad \forall l \ge i \ge 1, \end{aligned}$$

indeed $F_i/F_{i-1}$ is nilpotent and normal in $F_l/F_{i-1}$ hence $F_i/F_{i-1} \subseteq F(F_l/F_{i-1})$, and $F(F_l/F_{i-1})$ is nilpotent and characteristic in $F_l/F_{i-1}$, which is normal in $G/F_{i-1}$, so $F(F_l/F_{i-1})$ is normal in $G/F_{i-1}$ hence $F(F_l/F_{i-1}) \subseteq F(G/F_{i-1}) = F_i/F_{i-1}$. This implies in particular that $F_l/F_i$ has Fitting height $l-i$.

The following consequence of the solution of the k(GV) problem is proved in (Guralnick and Robinson 2006, Lemma 3, proof of (i)) (see also Knorr 1984).

Proposition 3

Let G be a group and let F be the Fitting subgroup of G. If G / F is nilpotent (that is, G has Fitting height 2) then $k(G) \le |F|$, so that

$$\begin{aligned} {{\mathrm{cp}}}(G) \le \frac{1}{|G:F|}. \end{aligned}$$

The following result shows that $S_4$ is a “maximal” non-supersolvable group in terms of $\alpha (G)$.

Theorem 6

Let G be a group. If G is not supersolvable then $\alpha (G) \le \alpha (S_4)$ with equality if and only if $G \cong {C_2}^n \times S_4$ for some non-negative integer n.

Proof

We prove that if $\alpha (G) \ge \alpha (S_4)$ and G is not supersolvable then G is isomorphic to ${C_2}^n \times S_4$ for some non-negative integer n. We have $\alpha (G) \ge \alpha (S_4) = 17/24 > 67/120$ so G is solvable by Theorem 4, so the Fitting subgroup F of G is non-trivial, and since G is not supersolvable $G \ne F$. Since $2/3 < 17/24$, Proposition 1 implies that G does not have non-trivial quotients of odd order. Also, since $17/24 > 1/2$ we have ${{\mathrm{cp}}}(G) \ge (2 \alpha (G)-1)^2 \ge 25/144$ by Lemma 2.

In the following discussion we will use Proposition 3, the inequality ${{\mathrm{cp}}}(G) \le {{\mathrm{cp}}}(N) \cdot {{\mathrm{cp}}}(G/N)$ for $N \unlhd G$ (see Guralnick and Robinson 2006, Lemma 1) and the obvious fact that the commuting probability is always at most 1. Let $F_i$ be the subgroups defined above and let h be the Fitting height of G. We distinguish three cases.

• 1. $h = 2$. In this case $G = F_2> F_1 > \{1\}$. We have that $G/F_1$ is nilpotent so $25/144 \le {{\mathrm{cp}}}(G) \le 1/|G:F_1|$ so $|G:F_1| \le 5$. However $|G:F_1| \not \in \{3,5\}$ because G does not have non-trivial quotients of odd order, so $G/F_1$ is one of $C_2$, $C_4$ and $C_2 \times C_2$.

• 2. $h = 3$. In this case $G = F_3> F_2> F_1 > \{1\}$. We have $25/144 \le {{\mathrm{cp}}}(G) \le {{\mathrm{cp}}}(F_2) \le 1/|F_2:F_1|$ and $25/144 \le {{\mathrm{cp}}}(G) \le {{\mathrm{cp}}}(G/F_1) \le 1/|G:F_2|$ so $|F_2:F_1| \le 5$ and $|G:F_2| \le 5$. Also $G/F_1$ is not a group of prime power order (because it is not nilpotent) and $|G:F_2|$ is not 3 or 5 because G does not have quotients of odd order. Therefore $G/F_1$ is a group of order 6, 10, 12 or 20, its Fitting subgroup has order at most 5 and $\alpha (G/F_1) \ge \alpha (G) \ge 17/24$. We deduce $G/F_1 \cong S_3$ by (The GAP Group 2018).

• 3. $h \ge 4$. In this case $G \ge F_4> F_3> F_2> F_1 > \{1\}$. We have

  $$\begin{aligned} \frac{25}{144} \le {{\mathrm{cp}}}(G) \le {{\mathrm{cp}}}(F_2) \cdot {{\mathrm{cp}}}(F_4/F_2) \le \frac{1}{|F_2:F_1|} \cdot \frac{1}{|F_4:F_3|} \end{aligned}$$

  so $|F_2:F_1| \cdot |F_4:F_3| \le 5$ implying that $|F_2:F_1| = |F_4:F_3| = 2$. But then $F(F_3/F_1) = F_2/F_1 \subseteq Z(F_3/F_1)$ implying $F_3/F_1=F_2/F_1$ (because the Fitting subgroup contains its own centralizer), a contradiction.

We deduce that G / F is one of $C_2$, $C_4$, $C_2 \times C_2$ and $S_3$.

Since G is not supersolvable there exists a maximal subgroup M of G whose index |G : M| is not a prime number (see Robinson 1996, 9.4.4). Let $M_G$ the normal core of M in G, that is, the intersection of the conjugates of M in G. Let $X := G/M_G$, $K := M/M_G$, so that $|G:M|=|X:K|$. Then $\alpha (S_4) \le \alpha (G) \le \alpha (X)$. This implies that if $X \cong S_4$ then the result follows from Theorem 1, so all we have to prove is that $X \cong S_4$. The subgroup $M/M_G$ of X is maximal and it has trivial normal core, so X is a primitive solvable group. We will make use of the known structural properties of primitive solvable groups, see for example (Doerk and Hawkes 1992, Section 15 of Chapter A). X is a semidirect product $X = V \rtimes K$ with $V={C_p}^n$, p a prime, V is the unique minimal normal subgroup of X and it equals the Fitting subgroup of X. Since $|V|=|X:K|$ is not a prime, $n \ge 2$, so $|X| > 6 \ge |G/F|$ hence $F \not \subseteq M_G$ so $FM_G/M_G$ is a non-trivial nilpotent normal subgroup of X so it equals V, hence

$$\begin{aligned} K \cong X/V = (G/M_G)/(FM_G/M_G) \cong G/FM_G \end{aligned}$$

is a quotient of G / F so K is one of $C_2$, $C_4$, $C_2 \times C_2$ and $S_3$.

In what follows we will use the known representation theory of small groups over the field with p elements. We will think of V as a vector space of dimension n over the field $\mathbb {F}_p$, irreducible when seen as a $\mathbb {F}_p[K]$-module.

Suppose p divides |K|. Observe that being not supersolvable, X is not a 2-group, hence K cannot be $C_2$, $C_4$ nor $C_2 \times C_2$, so $K \cong S_3$. The structure of the group algebras $\mathbb {F}_2[S_3]$ and $\mathbb {F}_3[S_3]$ implies that $n \ge 2$ forces $n=p=2$ and hence $X \cong S_4$.

Suppose now p does not divide |K|, so that |K| and |V| are coprime.

Suppose $\mathbb {F}_p$ is a splitting field for K. Since $n \ge 2$, the only possibility is $K \cong S_3$, $n=2$, and the action of K on V defining the group structure of $X = V \rtimes K$ is the following: $K \cong S_3$ permutes the coordinates of the vectors in the fully deleted module

$$\begin{aligned} V = \{(a,b,c) \in {\mathbb {F}_p}^3\ :\ a+b+c=0\}. \end{aligned}$$

The elements of order 2 in X are of the form (a, b, c)k with $k \in K$, $o(k)=2$ and the fixed coordinate is zero, so there are 3p of them. The elements of order 3 in X are of the form vk with $v \in V$ arbitrary and $k \in K$, $o(k)=3$, so there are $2p^2$ of them. The elements of order p in X are the non-trivial elements of V (being p coprime to $|K|=6$) so there are $p^2-1$ of them. The elements of order 2p in X are of the form vk with $v \in V$, $k \in K$, $o(k)=2$ and $o(vk) \ne 2$ so there are $3p^2-3p$ of them. The exponent of X is 6p and X has no elements of order 6, 3p or 6p. This implies that

$$\begin{aligned} c(X)&= \sum _{x \in X} \frac{1}{\varphi (o(x))} = 1+3p+\frac{2p^2}{2}+\frac{p^2-1}{p-1}+\frac{3p^2-3p}{p-1} = p^2+7p+2. \end{aligned}$$

Using $p \ge 5$ and $|X|=6p^2$ we deduce $\alpha (X) < 17/24$, a contradiction.

Suppose $\mathbb {F}_p$ is not a splitting field for K. Since $n \ge 2$, the only possibility is $K \cong C_4$, $n=2$, and the polynomial $t^2+1$ does not split modulo p, that is, $p \equiv 3 \mod 4$. Let x be a generator of the cyclic group K. We may interpret x as a matrix of order 4, so that by irreducibility the minimal polynomial of x is $t^2+1$ (the only irreducible factor of degree 2 of $t^4-1$). This implies that $x^2=-1$. Choosing a nonzero vector $v_1 \in V$ and $v_2:={v_1}^x$, since $v_1,v_2$ are linearly independent (otherwise $v_1$ would be an eigenvector for x contradicting irreducibility), and ${v_2}^x={v_1}^{x^2}=-v_1$, the matrix of x in the base $\{v_1,v_2\}$ is $\left( \begin{array}{cc} 0 &{} 1 \\ -1 &{} 0 \end{array} \right)$ acting by right multiplication. $X = V \rtimes K$ has $p^2-1$ elements of order p (the non-trivial elements of V, being p coprime to $|K|=4$), $p^2$ elements of order 2 (the elements of the form $vx^2$ with $v \in V$ arbitrary) and $2p^2$ elements of order 4 (the elements of the form vx or $vx^3$ with $v \in V$ arbitrary). The exponent of X is 4p and there are no elements of order 2p or 4p, therefore

$$\begin{aligned} c(X) = \sum _{x \in X} \frac{1}{\varphi (o(x))} = 1+p^2+\frac{2p^2}{2}+\frac{p^2-1}{p-1} = 2p^2+p+2. \end{aligned}$$

Using $p \ge 3$ and $|X|=4p^2$ we deduce $\alpha (X) < 17/24$, a contradiction. $\square$