1 Introduction

It is well known that a finite group G cannot be the union of two proper subgroups. In fact, an easy counting argument shows that such a union covers at most three quarters of the elements of G. It is equally well known that G is never a union of conjugates of a proper subgroup H. Cameron–Cohen [4] have shown more precisely that there are at least |H| elements outside such a union. On the other hand, it may happen that G is covered by \(n\ge 3\) arbitrary proper subgroups \(H_1,\ldots ,H_n\le G\). While many authors classified such groups for a given n (the interested reader is referred to the survey [2]), we are interested in the situation where \(H_1\cup \ldots \cup H_n\ne G\). We show that a portion of elements, depending only on n, lies outside this union. In fact, this holds more generally for union of cosets of subgroups. To the authors’ knowledge, this has apparently not been observed in the literature. In the first part of the paper, we prove more precisely that |G|/(2n!) elements lie outside such a coset union. In the second part, we investigate our conjecture that even \(|G|/2^n\) elements lie outside the union. For elementary abelian groups, we obtain in Theorem 6 the best possible bound of that kind by using a linear algebra approach due to Alon–Füredi [1]. For cyclic groups, our results are closely related to a number theoretical problem by Erdős [5] (the details are outlined at the end of the paper). We like to mention that there are other open conjectures on union of cosets such as the long-standing Herzog–Schönheim Conjecture [6, 8].

2 Main result

Theorem 1

For every positive integer n, there exists a constant \(\gamma _n<1\) with the following property: For every finite group G and every n subgroups \(H_1,\ldots ,H_n\le G\) and \(g_1,\ldots ,g_n\in G\) either \(g_1H_1\cup \ldots \cup g_nH_n=G\) or \(|g_1H_1\cup \ldots \cup g_nH_n|\le \gamma _n|G|\).

Proof

We argue by induction on n. For \(n=1\), the claim holds with \(\gamma _1=\frac{1}{2}\) by Lagrange’s Theorem. Now let \(n\ge 2\), \(H_1,\ldots ,H_n\le G\) and \(g_1,\ldots ,g_n\in G\) such that \(g_1H_1\cup \ldots \cup g_nH_n\ne G\). Let \(s_i:=|G:H_i|\) for \(i=1,\ldots ,n\). We may assume that \(s_1\le \ldots \le s_n\). Let \(\alpha _n\) be the smallest positive integer such that \(\gamma _{n-1}+\frac{1}{\alpha _n}<1\). If \(s_n\ge \alpha _n\), then induction yields

$$\begin{aligned} |g_1H_1\cup \ldots \cup g_nH_n|\le & {} |g_1H_1\cup \ldots \cup g_{n-1}H_{n-1}|+|g_nH_n|\\\le & {} \Bigl (\gamma _{n-1}+\frac{1}{s_n}\Bigr )|G| \le \Bigl (\gamma _{n-1}+\frac{1}{\alpha _n}\Bigr )|G|. \end{aligned}$$

Now let \(s_n\le \alpha _n\) and \(H:=H_1\cap \ldots \cap H_n\). Using Poincaré’s formula \(|G:H_i\cap H_j|\le |G:H_i||G:H_j|\) repeatedly, we get \(|G:H|\le s_1\ldots s_n\le \alpha _n^n\). Since \(g_1H_1\cup \ldots \cup g_nH_n\) is a union of H-cosets, it follows that

$$\begin{aligned} |g_1H_1\cup \ldots \cup g_nH_n|\le |G|-|H|\le \Bigl (1-\frac{1}{\alpha _n^n}\Bigr )|G|. \end{aligned}$$

Hence, the claim holds with

$$\begin{aligned} \gamma _n:=\max \Bigl \{\gamma _{n-1}+\frac{1}{\alpha _n},1-\frac{1}{\alpha _n^n}\Bigr \}<1. \end{aligned}$$

\(\square \)

The proof of Theorem 1 yields only a very crude bound on \(\gamma _n\). With some more effort, we can prove an effective bound as follows.

Proposition 2

Theorem 1 holds with \(\gamma _n=\frac{2n!-1}{2n!}\).

Proof

We reuse the notation from the proof of Theorem 1. We already know that the claim holds for \(n=1\). Thus, let \(n\ge 2\). If \(n+1\le s_1\le \ldots \le s_n\), then

$$\begin{aligned} |g_1H_1\cup \ldots \cup g_nH_n|\le |H_1|+\cdots +|H_n|\le \frac{n}{n+1}|G|\le \frac{2n!-1}{2n!}|G| \end{aligned}$$

as desired.

Now let \(s_1\le n\). Since G is the union of all cosets of \(H_1\), there exists a coset \(gH_1\) such that \(gH_1\nsubseteq g_1H_1\cup \ldots \cup g_nH_n\). Since \(|g_1H_1\cup \ldots \cup g_nH_n|=|g^{-1}(g_1H_1\cup \ldots \cup g_nH_n)|\), we may replace \(g_i\) by \(g_i':=g^{-1}g_i\) for \(i=1,\ldots ,n\). Then, \(H_1\nsubseteq g_1'H_1\cup \ldots \cup g_n'H_n\) and \(H_1\cap g_1'H_1=\varnothing \). It follows that

$$\begin{aligned} g_1'H_1\cup \ldots \cup g_n'H_n\subseteq (G\setminus H_1)\mathbin {\dot{\cup }} \bigcup _{i=2}^n(g_i'H_i\cap H_1). \end{aligned}$$

If \(g_i'H_i\cap H_1\ne \varnothing \), then \(g_i'H_i\cap H_1=h_i(H_i\cap H_1)\) for some \(h_i\in H_1\). By induction on n, we conclude that

$$\begin{aligned} \Bigl |(G\setminus H_1)\mathbin {\dot{\cup }}\bigcup _{i=2}^n(g_i'H_i\cap H_1)\Bigr |\le \frac{s_1-1}{s_1}|G|+\gamma _{n-1}|H_1|\le \frac{s_1+\gamma _{n-1}-1}{s_1}|G|. \end{aligned}$$

Since \(\gamma _{n-1}-1<0\), it follows that

$$\begin{aligned} \frac{s_1+\gamma _{n-1}-1}{s_1}\le \frac{n+\gamma _{n-1}-1}{n}=\frac{2n!+2(n-1)!-1-2(n-1)!}{2n!}=\gamma _n \end{aligned}$$

as desired. \(\square \)

In most cases, cosets can cover more elements than subgroups. For instance, if G is a p-group, then two distinct cosets of a maximal subgroup cover \(\frac{2}{p}|G|\) elements while two distinct maximal subgroups only cover \(\frac{2p-1}{p^2}|G|\) elements (see also Theorem 6).

In order to compute a lower bound on \(\gamma _n\), let us consider an elementary abelian 2-group \(G=\langle x_1,\ldots ,x_n\rangle \) rank n. Let \(H_i:=\langle x_j:j\ne i\rangle \). Then, \(H_1\cup \ldots \cup H_n=G\setminus \{x_1\ldots x_n\}\), and therefore, \(\gamma _n\ge (2^n-1)/2^n\) for all \(n\ge 1\). If we restrict ourselves to union of subgroups, we can show that this bound is indeed optimal for small n.

Proposition 3

For every finite group G and every set of subgroups \(H_1,\ldots ,H_n\le G\) with \(n\le 5\) either \(H_1\cup \ldots \cup H_n=G\) or \(|H_1\cup \ldots \cup H_n|\le \frac{2^n-1}{2^n}|G|\). Equality can only hold if \(|G:H_1\cap \ldots \cap H_n|=2^n\).

Proof

We may assume that \(n\ge 2\), \(H_1\cup \ldots \cup H_n\ne G\) and \(H_i\nsubseteq \bigcup _{j\ne i}H_j\) for \(i=1,\ldots ,n\). Let \(N:=\{1,\ldots ,n\}\) and \(H_I:=\bigcap _{i\in I}H_i\) for \(I\subseteq N\). Suppose first that \(L:=H_{N\setminus \{i\}}\nsubseteq H_i\) for some i, say \(i=1\). Let \(U:=G\setminus \bigcup _{i=2}^nH_i\). By induction on n, we have \(|U|\ge |G|/2^{n-1}\). Moreover, U is a union of L-cosets. If \(g\in G\) and \(x\in gL\cap H_1\), then

$$\begin{aligned} |gL\cap H_1|=|x(L\cap H_1)|=|L\cap H_1|=|H_N|. \end{aligned}$$

Hence, \(|U\cap H_1|\le \sum _{gL\subseteq U}|H_N|=\frac{|U|}{|L|}|H_N|\). It follows that

$$\begin{aligned} \Bigl |G\setminus \bigcup _{i=1}^nH_i\Bigr |=|U\setminus H_1|=|U|-|U\cap H_1|\ge |U|\Bigl (1-\frac{1}{|L:H_N|}\Bigr )\ge \frac{|G|}{2^n} \end{aligned}$$

as desired. Equality can only hold if \(|U|=|G|/2^{n-1}\) and \(|L:H_N|=2\). In this case, induction yields \(|G:L|=2^{n-1}\) and \(|G:H_N|=2^n\).

Hence, in the following we will assume that \(H_{N\setminus \{i\}}\subseteq H_i\) for \(i=1,\ldots ,n\). In particular, \(n\ge 3\). Since \(H_1\cup \ldots \cup H_n\) is a union of \(H_N\)-cosets, we may also assume that \(|G:H_N|>2^n\) as in the proof of Theorem 1. We need to show the strict inequality \(|H_1\cup \ldots \cup H_n|<\frac{2^n-1}{2^n}|G|\). Using

$$\begin{aligned} |H_1\cup \ldots \cup H_n|\le |H_1\cup \ldots \cup H_{n-1}|+|H_n|-|H_1\cap H_n|\le \Bigl (\gamma _{n-1}+\frac{1}{s_n}-\frac{1}{s_n^2}\Bigr )|G| \end{aligned}$$

and induction, the indices \(s_i:=|G:H_i|\) can be bounded. In particular, there are only finitely many choices. By using

$$\begin{aligned} |G:H_I|\ \bigm |\ |G:H_{I\cup J}|\le |H_{I\cap J}:H_I||G:H_J|=\frac{|G:H_I||G:H_J|}{|G:H_{I\cap J}|} \end{aligned}$$

for \(I,J\subseteq N\), we can enumerate all possible indices \(|G:H_I|\) for \(I\subseteq N\) by computer. The claim can then be checked with the exclusion-inclusion principle. Note that for \(n=3\) this becomes

$$\begin{aligned} |H_1\cup H_2\cup H_3|=\Bigl (\frac{1}{s_1}+\frac{1}{s_2}+\frac{1}{s_3}-\frac{2}{|G:H_N|}\Bigr )|G| \end{aligned}$$

where \(9\le |G:H_N|=|G:H_1\cap H_2|\le s_1s_2\) and \(s_1\le s_2\le s_3\). It is easy to see that this implies \(|H_1\cup H_2\cup H_3|\le \frac{7}{9}|G|\) with equality if and only if \(s_1=s_2=s_3=3\). For \(n=4\), we obtain similarly

$$\begin{aligned} |H_1\cup \ldots \cup H_4|\le \Bigl (\frac{1}{s_1}+\cdots +\frac{1}{s_4}-\frac{1}{s_1s_2}-\frac{1}{s_1s_3}-\ldots -\frac{1}{s_3s_4}+\frac{3}{|G:H_N|}\Bigr )|G| \end{aligned}$$

where \(|G:H_N|\ge 17\) (in fact, \(|G:H_N|\ge 18\) since \(|G:H_N|\) cannot be a prime). By induction, we get \(s_4\le 15\). The maximum for the estimate above is assumed for \(s_1=s_2=2\), \(s_3=s_4=15\) (checked by computer). This maximum is again strictly less than \(\frac{15}{16}|G|\).

Finally, let \(n=5\). Here we first estimate the union of four out of five subgroups. This leaves us with a short list of exceptional cases. In all those cases, there exist subgroups \(A,B,C,D\in \{H_1,\ldots ,H_5\}\) with the following indices

$$\begin{aligned} |G:A\cap B|&=4,&|G:C\cap D|&=|G:C||G:D|,\\ |G:A\cap B\cap C|&=2|G:C|,&|G:A\cap B\cap D|&=2|G:D|. \end{aligned}$$

From \(|G:A\cap B|=4\), we obtain \(|G:A|=|G:B|=2\) and \(A\cap B\unlhd G\) (the proof of Proposition 2 now already yields \(\gamma _5=31/32\), but we need a strict inequality here). In particular, \((A\cap B)C\le G\) with \(|(A\cap B)C:A\cap B|=|C:A\cap B\cap C|=2\). Thus, \(A,B,(A\cap B)C\) are the maximal subgroups of G containing \(A\cap B\), and therefore, \(G=A\cup B\cup (A\cap B)C\). From \(|G:A\cap B\cap D|=2|G:D|=|G:D\cap A|=|G:D\cap B|\) we obtain \(D\cap A=D\cap B\). It follows that

$$\begin{aligned} D=(D\cap A)\cup (D\cap B)\cup (D\cap (A\cap B)C)=(D\cap A)\cup (D\cap (A\cap B)C). \end{aligned}$$

Since D is not the union of two proper subgroups, we conclude that \(D\subseteq (A\cap B)C\). But also \(C\subseteq (A\cap B)C\). Now \(G=CD\subseteq (A\cap B)C\), because \(|G:C\cap D|=|G:C||G:D|\). Contradiction. \(\square \)

We remark that the following alternative procedure applies more generally to union of cosets. Note that G acts on \(\bigcup _{i=1}^nG/H_i\) by left multiplication. The kernel N of this action is contained in \(H_1\cap \ldots \cap H_n\). Since \(g_1H_1\cup \ldots \cup g_nH_n\) is the union of the cosets in \(g_1(H_1/N)\cup \ldots \cup g_n(H_n/N)\), we may replace G by G/N. Then G is isomorphic to a subgroup of a direct product of symmetric groups \(\prod _{i=1}^n S_{|G:H_i|}\). In principle, we can enumerate those subgroups by computer, but doing so becomes impractical when n is large.

3 Nilpotent groups

In order to extend Proposition 3 to other cases, we provide a reduction theorem for nilpotent groups. Let \(\delta _n(G)\) be the largest constant such that \(|G\setminus (g_1H_1\cup \ldots \cup g_nH_n)|\ge \delta _n(G)|G|\) whenever \(g_1H_1\cup \ldots \cup g_nH_n\ne G\). We wish to show that \(\delta _n(G)\ge 1/2^n\).

Lemma 4

Let \(n\ge 1\). Suppose that for every p-group P and every \(m\le n\) we have \(\delta _m(P)\ge 1/2^m\). Then, \(\delta _n(G)\ge 1/2^n\) for every nilpotent group G.

Proof

Let G be a nilpotent group. Let \(p_1,\ldots ,p_k\) be the distinct prime divisors of |G|. Let \(P_i:=\mathrm {O}_{p_i}(G)\) be the Sylow \(p_i\)-subgroup and \(Q_i:=\mathrm {O}_{p_i'}(G)\) its normal complement. Note that \(G=P_1\times \cdots \times P_k\). Let \(g_1H_1,\ldots ,g_nH_n\) be cosets of subgroups of G such that \(g_1H_1\cup \ldots \cup g_nH_n\ne G\). Suppose that \(|G:H_1|\) is divisible by \(p_i\) and \(p_j\) with \(i\ne j\). Let \(K:=H_1\mathrm {N}_{P_i}(H_1)\) and \(L:=H_1\mathrm {N}_{P_j}(H_1)\). Then, \(g_1H_1=g_1(K\cap L)=g_1K\cap g_1L\) and \(g_1K\cup g_2H_2\cup \ldots \cup g_nH_n\ne G\) or \(g_1L\cup g_2H_2\cup \ldots \cup g_nH_n\ne G\). Thus, we may replace \(H_1\) by K or L, respectively. Since every subgroup of G is subnormal, we may continue in this way until \(|G:H_1|\) is a prime power. We repeat this process with \(H_i\) for \(i=2,\ldots ,k\). Then, every \(H_i\) contains a unique \(Q_j\).

Let \(\mathcal {H}_i:=\{H_j:Q_i\subseteq H_j\}\) for \(i=1,\ldots ,k\). Then, \(\{H_1,\ldots ,H_n\}\) is the disjoint union of \(\mathcal {H}_1,\ldots ,\mathcal {H}_k\). In particular, \(n=|\mathcal {H}_1|+\cdots +|\mathcal {H}_k|\). Moreover, an element \((x_1,\ldots ,x_k)\in P_1\times \cdots \times P_k\) does not lie in \(g_1H_1\cup g_2H_2\cup \ldots \cup g_nH_n\) if and only if \(x_iQ_i\) does not lie in \(\bigcup _{H_j\in \mathcal {H}_i}g_j(H_j/Q_i)\) for \(i=1,\ldots ,k\). If we regard \(H_j/Q_i\) as subgroups of \(P_i\cong G/Q_i\), it follows that

$$\begin{aligned} |G\setminus (g_1H_1\cup \ldots \cup g_nH_n)|\ge \prod _{i=1}^k\delta _{|\mathcal {H}_i|}(P_i)|P_i|\ge \prod _{i=1}^k\frac{|P_i|}{2^{|\mathcal {H}_i|}}=\frac{1}{2^n}|G|. \end{aligned}$$

\(\square \)

Unfortunately, we are unable to prove \(\delta _n(P)\ge 1/2^n\) for p-groups in general. Nevertheless, we provide an optimal bound for elementary abelian p-groups by making use of combinatorial theorems of Alon–Füredi [1] (see also Theorem [9]). The following variant of the Schwartz–Zippel Lemma is an explicit version of [1, Theorem 5].

Lemma 5

Let p be a prime, and let \(\alpha \in \mathbb {F}_p[X_1,\ldots ,X_k]\) be a polynomial of total degree \(d=a+b(p-1)\) where \(0\le a\le p-2\). If \(\alpha \) does not vanish identically on \(\mathbb {F}_p^k\), then \(\alpha \) is non-zero on at least \(p^{k-b-1}(p-a)\) points of \(\mathbb {F}_p^k\). This bound is best possible for \(d\le k(p-1)\).

Proof

We argue by induction on k. Without loss of generality, we may assume that \(k>b\). If \(k=1\), then \(\alpha \) has at most \(d=a\) roots in \(\mathbb {F}_p\), so it is nonzero on at least \(p-a\) points. Now let \(k\ge 2\). By Fermat’s little theorem, \(x^p=x\) for all \(x\in \mathbb {F}_p\). Hence, we can reduce all powers of \(X_1\) such that the degree of \(\alpha \) in \(X_1\) is at most \(p-1\). This might decrease d, so the bound will be even stronger. For \(x\in \mathbb {F}_p\), let

$$\begin{aligned} \gamma _x:=\alpha (x,X_2,\ldots ,X_k)\in \mathbb {F}_p[X_2,\ldots ,X_k]. \end{aligned}$$

Let \(C\subseteq \mathbb {F}_p\) be the set of \(x\in \mathbb {F}_p\) such that \(\gamma _x\) does not vanish identically on \(\mathbb {F}_p^{k-1}\). By hypothesis, \(C\ne \varnothing \). Let \(p':=p-|C|\). Let

$$\begin{aligned} \alpha =\alpha _1X_1^{p-1}+\alpha _2X_1^{p-2}+\cdots +\alpha _p \end{aligned}$$

with \(\alpha _i\in \mathbb {F}_p[X_2,\ldots ,X_k]\) and \(\deg (\alpha _i)\le d-p+i\) for \(i=1,\ldots ,p\). We arrange the elements of \(\mathbb {F}_p^{k-1}\) in some fixed order, say \(\mathbb {F}_p^{k-1}=\{v_1,v_2,\ldots ,v_{p^{k-1}}\}\), and define \(\overline{\alpha _i}:=(\alpha _i(v_1),\alpha _i(v_2),\ldots ,\alpha _i(v_{p^{k-1}}))^\mathrm {t}\in \mathbb {F}_p^{p^{k-1}\times 1}\). For \(x\in \mathbb {F}_p\setminus C\), we obtain a linear equation \(x^{p-1}\overline{\alpha _1}+x^{p-2}\overline{\alpha _2}+\cdots +\overline{\alpha _p}=0\). The Vandermonde matrix \(A:=(x^i:i=0,\ldots ,p'-1,x\in \mathbb {F}_p\setminus C)\) is invertible and

$$\begin{aligned} (\overline{\alpha _p},\overline{\alpha _{p-1}},\ldots ,\overline{\alpha _{|C|+1}})A=-(x^{p-1}\overline{\alpha _1}+\cdots +x^{p'}\overline{\alpha _{|C|}}:x\in \mathbb {F}_p\setminus C). \end{aligned}$$

Therefore, we can express the vectors \(\overline{\alpha _{|C|+1}},\ldots ,\overline{\alpha _p}\) as linear combinations of \(\{x^{p-1}\overline{\alpha _1}+\cdots +x^{p'}\overline{\alpha _{|C|}}:x\in \mathbb {F}_p\setminus C\}\). Hence, we may replace each \(\alpha _i\) with \(|C|<i\le p\) by a linear combination of \(\alpha _1,\ldots ,\alpha _{|C|}\) without changing the values on \(\mathbb {F}_p^{k-1}\). Eventually, \(\deg (\alpha _i)\le d-p'\) for all i and \(\deg (\gamma _x)\le d-p'\) for \(x\in C\). By induction, \(\gamma _x\) is non-zero on at least \(p^{k-b'-2}(p-a')\) points of \(\mathbb {F}_p^{k-1}\) where \(d-p'=a'+b'(p-1)\) with \(0\le a'\le p-2\). Consequently, \(\alpha \) is nonzero on at least

$$\begin{aligned} |C|p^{k-b'-2}(p-a')=p^{k-b'-2}(p-a')(p-p') \end{aligned}$$

points of \(\mathbb {F}_p^k\).

Suppose first that \(p'\le a\). Then, \(a'=a-p'\) and \(b'=b\). It follows that

$$\begin{aligned} p^{k-b-2}(p-a+p')(p-p')\ge p^{k-b-2}(p-a+p')(p-a)\ge p^{k-b-1}(p-a) \end{aligned}$$

and we are done. Now let \(a<p'\le p-1\). Then, \(a'=a-p'+p-1\) and \(b'=b-1\). Since \((p'-a)(p-p')\ge p'-a\), we obtain \((p'-a+1)(p-p')\ge p-a\). This yields \(p^{k-b'-2}(p-a')(p-p')\ge p^{k-b-1}(p-a)\) as desired.

To see that the bound is best possible, just consider

$$\begin{aligned} \alpha =\prod _{i=1}^b(X_i^{p-1}-1)\prod _{j=1}^a(X_{b+1}-j) \end{aligned}$$

where j is interpreted as \(1+\cdots +1\in \mathbb {F}_p\) (j summands). \(\square \)

Theorem 6

Let G be an elementary abelian p-group, \(H_1,\ldots ,H_n\le G\) and \(g_1,\ldots ,g_n\in G\) such that \(g_1H_1\cup \ldots \cup g_nH_n\ne G\). Let \(n=a+b(p-1)\) where \(0\le a\le p-2\). Then,

$$\begin{aligned} |g_1H_1\cup \ldots \cup g_nH_n|\le \frac{p^{b+1}-p+a}{p^{b+1}}|G|\le \frac{2^n-1}{2^n}|G| \end{aligned}$$

and the first inequality is best possible.

Proof

We regard G as the \(\mathbb {F}_p\)-vector space \(\mathbb {F}_p^k\). Each coset \(g_iH_i\) is the set of solutions of a linear system \(A_ix=b_i\). By hypothesis, there exists \(x\in G\setminus (g_1H_1\cup \ldots \cup g_nH_n)\). For each i, we choose a row \(a_i\) of \(A_i\) such that \(a_ix\ne \beta _i\) where \(\beta _i\in \mathbb {F}_p\) is the corresponding entry of \(b_i\). Then, the polynomial

$$\begin{aligned} \alpha (X_1,\ldots ,X_k):=\prod _{i=1}^n(a_i(X_1,\ldots ,X_k)^\mathrm {t}-\beta _i)\in \mathbb {F}_p[X_1,\ldots ,X_k] \end{aligned}$$

of degree n does not vanish on x. By Lemma 5, \(\alpha \) is nonzero on at least \(p^{k-b-1}(p-a)=\frac{p-a}{p^{b+1}}|G|\) points of G. All these points lie outside of \(g_1H_1\cup \ldots \cup g_nH_n\). This implies the first inequality. For the second, we may assume that \(a=p-2\) and \(b+1=\frac{n-a}{p-1}+1=\frac{n+1}{p-1}\). It suffices to show that \(2^{n+1}\ge p^{b+1}\), i. e. \((n+1)\log _p(2)\ge \frac{n+1}{p-1}\). This is true since \(2^{p-1}\ge p\).

In order to show that the first inequality is optimal, we choose \(H_1=\ldots =H_{p-1}\) as a maximal subgroup of G and \(g_1,\ldots ,g_{p-1}\in G\) such that \(G\setminus H_1=g_1H_1\cup \ldots \cup g_{p-1}H_1\). Similarly, choose \(H_p=\ldots =H_{2p-1}\) as a maximal subgroup of \(H_1\) and \(g_p,\ldots ,g_{2p-1}\in H_1\) such that \(H_1\setminus H_p=g_pH_p\cup \ldots \cup g_{2p-1}H_p\) and so on. This will certainly yield the exact bound. \(\square \)

We remark that Theorem 6 extends to arbitrary finite p-groups as long as \(n\le 2p-2\). To see this, consider \(g_1H_1\cup \ldots \cup g_nH_n\ne G\) where G is a finite p-group. If all \(H_1,\ldots ,H_n\) are maximal subgroups of G, then, by the remark at the end of Sect. 2, we can go over to the elementary abelian group \(G/\Phi (G)\) where \(\Phi (G)\) is the Frattini subgroup of G. In this case, the claim follows from Theorem 6. Otherwise, we may assume that \(H_n\) is not maximal. Then, the claim follows by induction on n, because

$$\begin{aligned} |g_1H_1\cup \ldots \cup g_nH_n|\le |g_1H_1\cup \ldots \cup g_{n-1}H_{n-1}|+|H_n|\\ \le |g_1H_1\cup \ldots \cup g_{n-1}H_{n-1}|+\frac{1}{p^2}|G|. \end{aligned}$$

On a different note, we mention that the subgroup lattice of some (but not all) p-groups can be embedded into the subgroup lattice of an elementary abelian p-group (see [3]). For instance, Theorem 6 carries over to cyclic p-groups (here, two cosets are either disjoint or one lies in the other). By virtue of Lemma 4, the bound \(\delta _n(G)\ge 1/2^n\) now holds for all cyclic groups. The reviewer pointed out that this is connected to the following arithmetical theorem of Erdős [5]: Suppose that \(a_1+m_1\mathbb {Z},\ldots ,a_n+m_n\mathbb {Z}\) are pairwise disjoint residue classes such that \(0<m_1<\ldots <m_n\). We may consider these residue classes as cosets inside \(\mathbb {Z}/M\mathbb {Z}\) where \(M:={\text {lcm}}(m_1,\ldots ,m_n)\). Then, the union of these (disjoint) cosets is \(\frac{M}{m_1}+\cdots +\frac{M}{m_n}\). Erdős showed that \(\frac{M}{m_1}+\cdots +\frac{M}{m_n}\le \frac{2^n-1}{2^n}M\) which is precisely our bound.

On the basis of these examples, the following general conjecture seems reasonable.

Conjecture 7

The best possible bound in Theorem 1 is \(\gamma _n=(2^n-1)/2^n\) for all n.