1 Introduction

Throughout G will be a finite group and p a prime dividing its order |G|. By Sylow, the set \(\textrm{Syl}_p(G)\) of Sylow p-subgroups of G is nonempty, and G acts transitively on it via conjugation. Let \({\pi }={\pi }_p(G)\) denote the associated permutation character of G (Sylow character).

Fix \(P\in \textrm{Syl}_p(G)\). Then \(|P|=|G|_p\) equals the p-part of |G| and \(\textrm{Syl}_p(G)=P^G=\{P^g: g\in G\}\) (where \(P^g=g^{-1}Pg\)). Each P-orbit on \(\textrm{Syl}_p(G)\) distinct from \(\{P\}\) has size a proper power of p. Hence the Sylow p-number \(s_p(G)=|\textrm{Syl}_p(G)|\) has the form \(s_p(G)=1+kp\) for some integer \(k\ge 0\). The set \(G_p=\bigcup _{g\in G}P^g\) of all p-elements in G can be described as \(G_p=\{g\in G:\; g^{|P|}=1\}\). Hence, by a theorem of Frobenius [2], \(f_p(G)=|G_p|/ |P|\) is an integer, called the Frobenius p-ratio of G. The identity

$$\begin{aligned} f_p(G)={1\over |P|}\sum _{x\in P}{s_p(G)\over {\pi }(x)} \end{aligned}$$

is basic for our purposes (see Lemma 3 below). Obviously \(s_p(G)\) and \(f_p(G)\) remain unchanged when replacing G by \(G/O_p(G)\) where \(O_p(G)=\bigcap _{g\in G}P^g\) is the greatest normal p-subgroup of G. We have \(P=O_p(G)\) if and only \(s_p(G)=1\) resp. \(f_p(G)=1\).

We say that the Sylow p-subgroups of G are irredundant (for \(G_p\)) provided none can be omitted in order to get \(G_p\) as their union. Of course this is true if and only if the given \(P\in \textrm{Syl}_p(G)\) cannot be omitted. Thus we have irredundancy in this sense precisely when \(\widehat{P}=\bigcup _{Q\in \textrm{Syl}_p(G){\setminus } \{P\}}(P\cap Q)\) is a proper subset of P. So P is a T.I. subset of G just when \(|\widehat{P}|=1\) and, as a rule, \(f_p(G)\) is large when \(|\widehat{P}|\) is small. We call P strongly irredundant provided \(|\widehat{P}|\le |P|/p\).

Theorem 1

Suppose \(P\in \textrm{Syl}_p(G)\) is not normal in G, so that the Sylow number \(s_p(G)=1+kp\) with \(k\ge 1\). Then the Frobenius ratio \(f_p(G)=1+\ell (p-1)\) for some positive integer \(\ell < {kp \over p-1}\). We have \(\ell \ge k\) when P is strongly irredundant, where \(\ell >k\), unless \(|P/O_p(G)|=p\). Conversely, if \(\ell \ge k\), then P is irredundant.

Thus \(f_p(G)\ge p\) (as \(\ell \ge 1\)). We call k and \(\ell \) the parameters of \(s_p(G)\) and \(f_p(G)\), respectively. If \(|G/O_p(G)|_p=p^a\), then \(\ell \le {kp\over p-1}(1-{1\over p^a})\), with equality precisely when \(P/O_p(G)\) is a T.I. subset of \(G/O_p(G)\). Observe that P is strongly irredundant when \(P/O_p(G)\) is cyclic or, more generally, when \(\widehat{P}=\bigcup _{x\in P}(P\cap Q)^x\) for some \(Q\in \textrm{Syl}_p(G)\). To see this, use that cyclic p-groups are uniserial, and that the normal closure of a proper subgroup of a finite p-group is a proper subgroup.

There exist groups G where \(s_p(G)>|G_p|\), and then at least one Sylow p-subgroup of G is redundant for \(G_p\), and we have \(\ell <k\) (see the example in Sect. 2). One has \((1+k(p-1))^p>(1+kp)^{p-1}\) by binomial expansion. Hence \(f_p(G) > s_p(G)^{p-1\over p}\) when \(\ell \ge k\ge 1\), and it is conjectured that this inequality holds in general. Indeed, Gheri [3] has confirmed this in the case where G is p-solvable, and has reduced this problem to almost simple groups. The authors have verified the conjecture for various such groups of small orders. It holds when G is a group of Lie type of characteristic p; use that then \(O_p(G)=1\) and that \(f_p(G)=|G|_p\) by a celebrated theorem of Steinberg [9].

Let \(r_p(G)\) denote the number of P-orbits on \(\textrm{Syl}_p(G)\), which evidently does not depend on the chosen \(P\in \textrm{Syl}_p(G)\). If \(s_p(G)=1+kp\) with \(k\ge 1\), then \(2\le r_p(G)\le 1+k\). We have \(r_p(G)=2\) if and only if P is transitive on the (nonempty) set \(\textrm{Syl}_p(G)\setminus \{P\}\), which happens when G is a Zassenhaus group of suitable degree. Also, \(r_p(G)=1+k\) if and only if each P-orbit on \(\textrm{Syl}_p(G)\setminus \{P\}\) has size p.

Theorem 2

Let \(s_p(G)=1+kp\) with \(k\ge 1\). Then \(f_p(G)>{s_p(G)\over r_p(G)}\). The Sylow p-subgroup P of G is strongly irredundant for \(G_p\) in the extreme cases \(r_p(G)=2\) and \(r_p(G)=1+k\).

In particular, we have strong irredundancy when \(k\le p\) since then \(r_p(G)=2\) or \(r_p(G)=1+k\).

Theorem 3

Let \(s_p(G)=1+kp\) with \(k\ge 2\). Then \(f_p(G)=1+\ell (p-1)\) with \(\ell \ge 2\), and the Sylow p-subgroup P of G is irredundant for \(G_p\) if \(s_p(G)\le p(f_p(G)+2)\).

Thus \(f_p(G)\ge 2p-1\) when \(s_p(G)\ge 1+2p\), and we certainly have irredundancy when \(s_p(G)\le p(2p+1)\).

2 The Sylow character

We keep the assumptions and notations as introduced above. Fix \(P\in \textrm{Syl}_p(G)\). We assume that P is not normal in G so that the Sylow number \(s_p(G)=1+kp\) with \(k\ge 1\). Then the Frobenius ratio \(f_p(G)>1\) as well. Recall that the Sylow character \({\pi }={\pi }_p(G)\) for the prime p is the permutation character of G in its action on \(\textrm{Syl}_p(G)=P^G\) (counting fixed points). For \(x\in G\), let \(x^G=\{x^g:\, g\in G \}\) denote the conjugacy class of x in G (so that \(|x^G|=|G:C_G(x)|\)).

Lemma 1

Let \(N=N_G(P)\) be the normalizer of P in G, and let \(\biguplus _{i=1}^rNg_iP\) be the decomposition of G into double cosets mod (N, P), say with \(g_1=1\). Then \(r=r_p(G)\) is the number of P-orbits on \(\textrm{Syl}_p(G)\) and \(s_p(G)=|G:N| =1+\sum _{i=2}^r|P:(P\cap P^{g_i})|\). In fact, the assignments \(P^g\mapsto Ng\) for \(g\in G\) define an isomorphism from the G-set \(\textrm{Syl}_p(G)\) onto the G-set \(G:N=\{Ng: g\in G\}\) (with G acting via multiplication from the right), and the P-sets \(P:(P\cap P^{g_i})\), \(i>1\), may be identified with the P-orbits on \(\textrm{Syl}_p(G)\setminus \{P\}\).

Proof

It is well known (and easily verified) that \(P^G\) is G-isomorphic to G : N (see Aufg. 23 on page 32 in Huppert [5]). In particular, \({\pi }\) is afforded by the G-set G : N, and is induced from the 1-character of N. So the remainder is nothing but Mackey decomposition (see Satz V.16.9 in [5]). The intersections \(P\cap P^{g_i}=P\cap N^{g_i}\) are point stabilizers for the corresponding P-orbits, and are determined up to P-conjugacy. \(\square \)

Note that \(N=N_G(P)\) is an abnormal subgroup of G [5, VI.11.18]; since \(N_G(N)=N\), one also knows that \(Ng\mapsto N^g\) for \(g\in G\) defines an isomorphism from G : N onto \(N^G\). Of course \(\textrm{Ker}({\pi })=\bigcap _{g\in G}N^g\), and \({\pi }(x)\ge 1\) if and only if \(x\in \bigcup _{g\in G}N^g\). Also \(O_p(G)\) is the kernel of P in its action on \(\textrm{Syl}_p(G)\), and \(\textrm{Ker}({\pi })/O_p(G)\) is a \(p'\)-group.

It is an interesting open question whether \(s_p(G)=|G:N_G(P)|\) can be read off from the character table of G (Navarro). Can one then obtain in addition \({\pi }\) as sum of the irreducible characters of G?

Lemma 2

Let \(x\in P\) and \(N=N_G(P)\). We have \({\pi }(x)=s_p(G)\cdot {|x^G\cap N|\over |x^G|}\) and \(x^G\cap N=x^G\cap P\). Moreover \({\pi }(x)\) is the number of Sylow p-subgroups of G containing \(\;x\) (including P), and \({\pi }(x)=1+k_xp\) for some nonnegative integer \(k_x\le k\). If H is a normal subgroup of G, then \(s_p(G)=s_p(PH)\cdot s_p(G/H)\) and \({\pi }(x)={\pi }_p(PH)(x)\cdot {\pi }_p(G/H)(Hx)\).

Proof

The first statement is well known (see Aufg. 24 on page 32 of [5]). The p-element \(x\in P\) normalizes some Sylow p-subgroup Q of G if and only if \(x\in Q\) (as Q is a p-subgroup of G of largest possible order, and \(Q\langle x \rangle \) is a p-subgroup of G). Moreover \(x^G\cap N=x^G\cap P\). We also have \({\pi }(x)\equiv 1\,(\textrm{mod}\,p)\) since each \(\langle x\rangle \)-orbit on \(\textrm{Syl}_p(G)\) has size a power of p, and since \(|\textrm{Syl}_p(G)|=s_p(G)=1+kp\). The last statement is due to Gheri [3, Lemma 3.6]. Use here that PH/H is a Sylow p-subgroup of G/H with normalizer NH/H, and consider the partition of the set of Sylow p-subgroups of G containing x (having cardinality \({\pi }(x)\)) where P and Q belong to the same part if \(PH=QH\). Since the Sylow p-subgroups of G/H are conjugate, there are \({\pi }_p(G/H)(Hx)\) parts with each part having cardinality \({\pi }_p(PH)(x)\). \(\square \)

Lemma 3

We have \(f_p(G)={1\over |P|}\sum _{x\in P} {s_p(G)\over {\pi }(x)}\).

Proof

Obviously \(G_p=\bigcup _{x\in P}x^G\). In this union, for any \(x\in P\), the conjugacy class \(x^G\) occurs with multiplicity \(|x^G\cap P|\). Thus \(|G_p|=\sum _{x\in P} {|x^G|\over |x^G\cap P|}\). Apply the preceding lemma. \(\square \)

In view of Lemma 3, and adopting an idea of Gheri [3], we define

$$\begin{aligned} f_p^*(G)=s_p(G)/\bigl (\prod _{x\in P}{\pi }(x)\bigr )^{1\over |P|}. \end{aligned}$$

Of course this again is independent of the chosen particular Sylow p-subgroup of G. By the arithmetic–geometric mean inequality, we have \(f_p(G)\ge f_p^*(G)\), and this inequality is proper since \(s_p(G)>1\), (and \({\pi }(x)\ne {\pi }(1)\) for some \(x\in P\)). Clearly \(s_p(G)\), \(r_p(G)\), \(f_p(G)\), and \(f_p^*(G)\) remain unchanged when replacing G by \(G/O_p(G)\) (or by \(O^{p'}(G)\)). Gheri [3] conjectured that always \(f_p^*(G)\ge s_p(G)^{p-1\over p}\), and confirmed this when G is p-solvable. By Lemma 2, we know that \(s_p(G)=s_p(PH)\cdot s_p(G/H)\) and \(f_p^*(G)=f_p^*(PH)\cdot f_p^*(G/H)\) for any normal subgroup H of G. Arguing by induction on |G|, one thus is readily reduced to the case where \(G=PH\) for some minimal normal subgroup H of G. Here H will be a \(p'\)-group when G is p-solvable, whence G p-nilpotent. This will be treated in Lemma 6 below.

Lemma 4

Let \(\widehat{P}=\bigcup _{Q\in \textrm{Syl}_p(G)\setminus \{P\}}(P\cap Q)\) be as introduced above. Then \(\widehat{P}\) is the subset of P consisting just of those \(x\in P\) for which \({\pi }(x)>1\); in other words, \({\pi }(x)=1\) if and only if \(x\in P\setminus \widehat{P}\).

Proof

This is immediate from Lemma 2. \(\square \)

Lemma 5

Suppose that \(O_p(G)=1\), and let \(1\ne x\in P\). Then \({\pi }(x)\le {s_p(G)\over 1+p}\) and hence \(k_x\le {k-1\over 1+p}\) (where \(s_p(G)=1+kp\) and \({\pi }(x)=1+k_xp\) are as above).

Proof

We use an idea given in the proof of Proposition 2.1 in [4]. This proposition might be incorrect as it stands, its proof being not conclusive as the sentence following statement \((*)\) can only be substantiated when the p-element considered has order p.

We have \({\pi }(x)\le {\pi }(y)\) if \(\langle y\rangle \subseteq \langle x\rangle \). So we may let x have order p. We assert that there is \(g\in G\) such that the sets of Sylow p-subgroups of G containing x and \(x^{gx^i}\) for \(i=0, \ldots , p-1\) are pairwise disjoint. Then we have \(1+p\) such sets, each of cardinality \({\pi }(x)\) as \({\pi }\) is a class function on G. Hence then \({\pi }(x)(1+p)\le s_p(G)\), as desired.

Assume the assertion is false. Then, for every \(g\in G\), there are indices \(i=i(g)\ne j(g)=j\) such that \(\langle x, x^g\rangle =\langle x, x^{gx^i}\rangle ^{x^{-i}}\) or \(\langle x, x^{gx^{j-i}g^{-1}}\rangle =\langle x^{gx^i}, x^{gx^j}\rangle ^{x^{-i}g^{-1}}\) are p-subgroups of G. In the latter case, \(j-i\) is not divisible by p. Application of a variant of the Baer–Wielandt theorem now gives us that \(\langle x\rangle \) is subnormal in G, whence \(x\in O_p(G)\) in contrast to our hypothesis.

The variant we mean is as follows. Let \(A=\langle x \rangle \) be a cyclic p-subgroup of G. Suppose that for every \(g\in G\), the join \(\langle x, x^{gyg^{-1}}\rangle \) is a p-group where either \(y=g\) or \(y=x^i\) for some integer \(i=i(g)\) not divisible by p. Then we claim that A is subnormal in G. Arguing by induction on |G|, and assuming that the claim is false, by Wielandt’s Zipper Lemma (originally proved in [10, Satz 5]), A is contained (and subnormal) in a unique maximal subgroup \(M=N_G(M)\) of G. For any \(g\in G\), this forces that either \(A^g\subseteq M\) (\(y=g\)), \(A\subseteq M^{g^{-1},}\) and \(g^{-1}\in M\), or that \(A^{gx^ig^{-1}}\subseteq M\), \((gx^ig^{-1})^{-1}\in M\), \(A=\langle x^{-i}\rangle \subseteq M^g\), and \(g\in M\). Thus, in each case, we get the contradiction \(G\subseteq M\). \(\square \)

Lemma 6

Suppose that \(G=PH\) is p-nilpotent, with normal p-complement H. Then \(s_p(G)=|H:C_H(P)|\) and \({\pi }(x)=|C_H(x):C_H(P)|\) for any \(x\in P\). Moreover

$$\begin{aligned} \bigl (\prod _{x\in P}{\pi }(x)\bigr )^{1\over |P|}\le s_p(G)^{1\over p}, \end{aligned}$$

where we have equality if and only if \(C_H(x^p)=H\) for every \(x\in P\). In particular, equality holds if P has exponent p, and then \(f_p^*(G)=s_p(G)^{p-1\over p}\).

Proof

We have \(|G:N_G(P)|=|H:C_H(P)|\) and \(x^G\cap P=x^P\) for \(x\in P\) [5, Satz IV.4.9]. From [5, Satz I.8.6], it follows that \(C_G(x)H=C_P(x)H\), whence \(|G:C_G(x)|\) equals \(|P:C_P(x)|\cdot |H:C_H(x)|\). Use finally that \(|C_H(P)|^{|P|}=\Bigl (\prod _{x\in P}{|C_H(x)|\over |C_H(x^p)|^{1\over p}}\Bigr )^{p\over p-1}\) by Theorem A in Navarro–Rizo [8]. \(\square \)

Example

For odd p the first example where a Sylow p-subgroup is redundant for \(G_p\) has been given by Derek Holt, namely for \(G=\textrm{PSL}_3(7)\) and \(p=3\). This can be found in a post given by Mikko Korhonen in MathematicsStackExchange (2013). There one also finds the following example by Jack Schmidt (see also Example 2.5 in [3]):

Let P be elementary abelian of order \(p^a\) (\(a\ge 2\)). Let \(q\equiv 1\,(\textrm{mod}\,p)\) be a prime and let H be an elementary abelian q-group of rank \({p^a-1 \over p-1}\), which is the number of maximal subgroups \(M_i\) of P. There is a faithful action of P on H such that \(H/C_H(M_i)\) has order q for each i (and \(\bigcap _iC_H(M_i)=1\)). The corresponding semidirect product \(G=H\cdot P\) is a p-nilpotent group satisfying \(C_H(P)=1\) and \(|C_H(x)|=q^{p^{a-1}-1\over p-1}\) for any \(x\ne 1\) in P, because there are just \({p^{a-1}-1\over p-1}\) maximal subgroups of P containing x. Hence \(s_p(G)=|H|\) and \({\pi }(x)=q^{p^{a-1}-1\over p-1}\) by Lemma 6. We infer that P is redundant for \(G_p\) and that \(f_p^*(G)=s_p(G)^{p-1\over p}\). Moreover \(|G_p|=1+(p^a-1)q^{p^{a -1}}\). We have \(q^{p^a-1\over p-1}>1+(p^a-1)q^{p^{a -1}}\) provided q is chosen large enough compared with \(p^a\) (which is possible by Dirichlet’s prime number theorem). In this case \(s_p(G)>|G_p|\).

3 Proof of Theorem 1

Fix \(P\in \textrm{Syl}_p(G)\), and let \(O_p(G)=1\) (without loss). Write \(|P|=p^a\) and \(s_p(G)=1+kp\) (\(a\ge 1\), \(k\ge 1\)). Let also \({\pi }={\pi }_p(G)\). Two nontrivial p-elements x, y of G are said to be related provided \(\langle x\rangle =\langle y \rangle \). This is an equivalence relation on \(G_p^\sharp =G_p\setminus \{1\}\), each class having cardinality divisible by \(p-1\). Thus \(|G_p^\sharp |=p^af_p(G)-1=(p^a-1)f_p(G)+(f_p(G)-1)\) is divisible by \(p-1\). Since \(p-1\) is a divisor of \(p^a-1\), we get that \(f_p(G)-1=\ell (p-1)\) for some integer \(\ell \), as required. Clearly \(\ell \ge 1\) (as \(k\ge 1\)). From Lemmas 3 and 2, it follows that \(f_p(G)<s_p(G)\), whence \(\ell < {kp\over p-1}\).

Let \(\widehat{P}=\{x\in P: {\pi }(x)>1\}\) (Lemma 4). If P is a T.I. subset of \(G=G/O_p(G)\), then \(|\widehat{P}|=1\) and \(f_p(G)={1\over p^a}\bigl (p^a(1+kp)-kp\bigr )=1+kp(1-{1\over p^a})=s_p(G)-{kp\over p^a}\) takes its greatest value (in terms of \(s_p(G)\) and |P|). Here \(p^{a-1}\) is a divisor of k (Frobenius). Let \(|\widehat{P}|>1\) in what follows, which forces that \(a\ge 2\). Using that \({\pi }(x)<s_p(G)\) for \(1\ne x\in \widehat{P}\) and \({\pi }(x)=1\) for \(x\in P\setminus \widehat{P}\) from Lemma 3, we get that

$$\begin{aligned} f_p(G) > {1\over |P|}\bigl (|\widehat{P}|+(|P|-|\widehat{P}|) (1+kp)\bigr ). \end{aligned}$$

If P is strongly irredundant, then \(|\widehat{P}|\le |P|/p=p^{a-1}\) (by definition), and we obtain that \(f_p(G)>1+kp-{kp |\widehat{P}|\over |P|}\ge 1+k(p-1)\).

Assume that \(f_p(G)\ge 1+k(p-1)\) but that P is redundant. Then \({\pi }(x)\ge 1+p\) for each \(x\in P\) by Lemmas 4 and 2, whence by Lemma 3,

$$\begin{aligned} 1+k(p-1) \le f_p(G)\le {1+kp\over 1+p}. \end{aligned}$$

It follows that \(k\bigl (p^2-p-1\bigr )\le -p\), which is impossible as \(k>0\) and \(p^2-p-1>0\).

Remark

Arguing as above, one gets that \(|\widehat{P}|=1+m(p-1)\) for some integer \(m\ge 0\) (even \(|\widehat{P}|/|\widehat{P}|_p\) is of this form). This could be helpful in order to check whether P is strongly irredundant. Of course there are groups G where a Sylow p-subgroup is irredundant but not strongly irredundant, and where the parameters \(\ell <k\). (For \(p=2\), such an example is provided by the symmetric group \(G=S_5\), where \(s_2(G)=15\) (\(k=7\)) and \(f_2(G)=7\) (\(\ell =6\)), and where \({\pi }(x)=1\) for any 4-cycle \(x\in G\).)

4 Proof of Theorem 2

Without loss let again \(O_p(G)=1\), and let \(s_p(G)=1+kp\) with \(k\ge 1\). Fix \(P\in \textrm{Syl}_p(G)\) and let \({\pi }={\pi }_p(G)\) be the Sylow character.

By the Cauchy–Frobenius–Burnside fixed point formula [5, Satz V.13.4] \(r_p(G)={1\over |P|}\sum _{x\in P}{\pi }(x)\), which is greater than \(\bigl (\prod _{x\in P}{\pi }(x)\bigr )^{1\over |P|}\) by the arithmetic–geometric mean inequality. Consequently

$$\begin{aligned} f_p(G)>f_p^*(G)>{s_p(G)\over r_p(G)}, \end{aligned}$$

as desired. Clearly \(2\le r_p(G)\le 1+k\), and \(r_p(G)=2\) if and only if P is transitive on \(\textrm{Syl}_p(G)\setminus \{P\}\), in which case \(\widehat{P}=\bigcup _{x\in P}(P\cap Q)^x\) for any \(Q\ne P\) in \(\textrm{Syl}_p(G)\).

Suppose that \(r_p(G)=1+k\). Then each P-orbit on \(\textrm{Syl}_p(G)\) different from \(\{P\}\) has size p. It follows that \(P=P/O_p(G)\) is elementary abelian (as it acts faithfully on \(\textrm{Syl}_p(G){\setminus } \{P\}\)). By Brodkey [1] there is \(Q\in \textrm{Syl}_p(G)\) such that \(P\cap Q=1\). But \(P\cap Q\) is a point stabilizer for the P-action on \(\textrm{Syl}_p(G)\) and \(|P:(P\cap Q)|=p\) is the size of a P-orbit. Thus \(|P|=p\) and, therefore, P is a T.I. subset of G. It follows that \(|G_p|=(1+kp)p-kp\) and \(f_p(G)=1+k(p-1)\). (In general there is no \(Q\in \textrm{Syl}_p(G)\) such that \(P\cap Q=O_p(G)\); e.g. see Ito [6].)

Remark

Let \(D=P\cap Q\) be a Sylow intersection of largest order (\(P\ne Q\in \textrm{Syl}_p(G)\)). Let \(|P:D|=p^b\) and \(|N_P(D):D|=p^c\) (\(1\le c\le b\)). Then the contribution to \(f_p(G)\) resulting from \(P\cup \bigcup _{x\in N_P(D)}Q^x\) is just \(1+p^c-p^{c-b}\), because the \(1+p^c\) Sylow subgroups involved intersect pairwise in D. Hence \(f_p(G)>p\) except possibly when \(b=c=1\).

5 Proof of Theorem 3

Let again \(O_p(G)=1\) (without loss), and let \(s_p(G)=1+kp\) with \(k\ge 2\). Fix also \(P\in \textrm{Syl}_p(G)\), and let \(|P|=p^a\). By Theorem 1, we know that \(f_p(G)=1+\ell (p-1)\) for some integer \(\ell \ge 1\). As usual, we let \({\pi }={\pi }_p(G)\).

We first show that the parameter \(\ell \ge 2\). Otherwise \(\ell =1\) and \(f_p(G)=p\), and we must have \(a\ge 2\) (since otherwise P is a T.I. subset of G). From Lemma 5 we get that \({\pi }(x)\le {s_p(G)\over 1+p}\) for each \(x\ne 1\) in P. Applying Lemma 3 we obtain that

$$\begin{aligned} p=f_p(G)\ge {1\over p^a}+{p^a-1\over p^a}(1+p)=1+p-{1\over p^{a-1}}>p, \end{aligned}$$

which is the desired contradiction. In particular, \(f_p(G)\ge 1+2(p-1)=2p-1\).

(A different approach can be given using the remark following Theorem 2, which in turn is in the spirit of [7, p. 80].)

Now let \(s_p(G)\le p(f_p(G)+2)\) or, equivalently, let \(k<f_p(G)+2\). We assert that then P is irredundant. Assume the assertion is false. It follows from Theorem 2 that we must have \(k\ge 1+p\), for otherwise P were even strongly irredundant. From Lemmas 4 and 2, we get that \({\pi }(x)\ge 1+p\), for each \(x\in P\). Using that \({\pi }(1)=1+kp>1+p\) therefore Lemma 3 shows that

$$\begin{aligned} 2p-1\le f_p(G)<{1+kp\over 1+p}=k-{k-1\over 1+p}\le k-{p\over 1+p}. \end{aligned}$$

Consequently \(k>2p-1+{p\over 1+p}\), whence \(k\ge 2p\ge p+2\). But this implies that \(f_p(G)<k-{k-1\over 1+p}\le k-1\) and so \(k\ge f_p(G)+2\), against our hypothesis.