Every Direct Power of A5 Satisfies the Herzog–Schönheim Conjecture
One bound on the index monoid settles every power at once, where the published route diverges
R003 · Novelty unconfirmed
The setting
A coset partition of a finite group \(G\) is an exact cover
\[G = H_1x_1 \;\sqcup\; H_2x_2 \;\sqcup\; \cdots \;\sqcup\; H_kx_k\]
by pairwise disjoint cosets of subgroups \(H_i \le G\). The index \([G:H]\) of a subgroup is the number of cosets it has; write \(I(G)\) for the set of distinct indices of subgroups of \(G\).
The Herzog–Schönheim conjecture, proposed in 1974, says that such a partition can never have all its indices pairwise distinct: whenever a finite group is covered exactly by \(k \ge 2\) cosets, two of those cosets must come from subgroups of the same index. A group with this property is called HS. The conjecture is still open. It is known for supersolvable groups, for groups of order below 1440 [3], and — in work of Garonzi and Margolis [1] — for all finite simple groups and all symmetric groups. Groups of order \(p^aq^b\) are known as well. Direct products of non-abelian simple groups are the case the published methods cannot reach, including the smallest interesting one, \(A_5 \times A_5\) of order 3600. The conjecture is listed as open at [4].
Everything below turns on one criterion, classical and stated as Lemma 2.1(1) of [1]. Write
\[\mathcal J(G) \;=\; \sum_{m \in I(G)} \frac{1}{m}\]
for the sum of the reciprocals of the distinct indices, counting \(m = 1\) for \(G\) itself. If \(\mathcal J(G) < 2\), then \(G\) is HS. The reason is a counting identity: in any coset partition the block sizes sum to \(|G|\), so
\[\sum_{i=1}^{k} \frac{1}{[G:H_i]} = 1 .\]
If all the indices were distinct and greater than \(1\), that sum would be at most \(\sum_{m \in I(G),\, m>1} 1/m = \mathcal J(G) - 1 < 1\) — impossible. The summand \(1/[G:G] = 1\) is exactly what the criterion compares against.
What is new, and what is not. The criterion \(\mathcal J(G) < 2 \Rightarrow\) HS is not new: it is classical, restated as Lemma 2.1(1) of [1]. What is new is the bound used to meet it: a reciprocal mass taken over the multiplicative monoid generated by the indices of the composition factors, which is uniform in the number of factors rather than growing with it. That is what makes direct powers accessible. The step is elementary — a published criterion applied to a bound that does not depend on \(r\) — and no new machinery is introduced; what it settles is whether such a bound exists at all.
The literature on powers. Garonzi and Margolis settle simple and symmetric groups [1], and their Remark 5.1 states that their methods are not sufficient for direct products of non-abelian simple groups, because the group \(H_n = A_{p_1} \times \cdots \times A_{p_n}\) acquires maximal subgroups of index \(p_1, \dots, p_n\), making the index sum unbounded. Searches across arXiv, zbMATH Open, OpenAlex, Semantic Scholar and the open web found no result for direct powers of a fixed simple group. MathSciNet and zbMATH full text were not reachable, so this is absence of evidence, not proof of novelty — hence the label on this page.
Not an algorithm. Nothing here factors, tests primality, or decides anything quickly. It is a counting bound on indices, and its content is that a single finite computation about \(A_5\) decides a question about \(A_5^r\) for every \(r\) at once.
The index monoid
For a set \(\mathcal S\) of finite simple groups, let \(M(\mathcal S)\) be the multiplicative monoid generated by the indices of the groups in \(\mathcal S\):
\[M(\mathcal S) \;=\; \Big\langle \; \bigcup_{S \in \mathcal S} I(S) \;\Big\rangle_{\text{multiplication}},\]
so \(1 \in M(\mathcal S)\), and define the reciprocal mass
\[\Lambda(\mathcal S) \;=\; \sum_{m \in M(\mathcal S),\; m > 1} \frac{1}{m} .\]
Two elementary facts make this the right object. The first controls how indices behave under extensions; it is the identity \(|G| = |N|\,|G/N|\) together with \(|H| = |H \cap N| \cdot |HN/N|\) for a subgroup \(H \le G\):
\[N \triangleleft G \quad\Longrightarrow\quad I(G) \;\subseteq\; I(N)\cdot I(G/N).\]
Iterating along a composition series \(1 = G_0 \triangleleft \cdots \triangleleft G_r = G\) gives \(I(G) \subseteq I(G_1/G_0)\cdots I(G_r/G_{r-1})\). The second is that a reciprocal sum over a subset of a monoid cannot exceed the sum over the whole monoid. Together they give the result.
Theorem 1 (The index-monoid theorem) Let \(\mathcal S\) be a finite set of finite simple groups, and suppose
\[\Lambda(\mathcal S) \;=\; \sum_{m \in M(\mathcal S),\, m>1} \frac{1}{m} \;<\; 1 .\]
Then every finite group whose composition factors lie in \(\mathcal S \cup \{C_p : p \in M(\mathcal S)\}\) satisfies the Herzog–Schönheim conjecture.
Corollary 1 (A5 and its powers) The monoid generated by the indices of \(A_5\) is \(M_5 = \langle 5,6,10,12,15,20 \rangle\), because
\[I(A_5) = \{1,\,5,\,6,\,10,\,12,\,15,\,20,\,30,\,60\}\]
and \(30 = 5\cdot 6\), \(60 = 5 \cdot 12\). Its reciprocal mass is the rational number
\[\sum_{m \in M_5} \frac{1}{m} \;=\; \frac{11463}{5852} \;=\; 1.95881749829\ldots \;<\; 2, \qquad\text{so}\qquad \Lambda(\{A_5\}) = \frac{5611}{5852} = 0.95881749829\ldots < 1 .\]
Consequently every finite group all of whose composition factors are isomorphic to \(A_5\) or \(C_5\) satisfies the conjecture — in particular
\[A_5^r \ \text{is HS for every } r \ge 1,\]
as are the groups \(A_5^r \times C_5^s\), and all of their subgroups and quotients.
The proof
Indices of a group lie in the monoid of its composition factors. For a composition series \(1 = G_0 \triangleleft \cdots \triangleleft G_r = G\), the extension identity gives \(I(G) \subseteq \prod_j I(G_j/G_{j-1})\). Each factor is either one of the groups in \(\mathcal S\) — contributing indices inside \(M(\mathcal S)\) — or a cyclic group \(C_p\) of prime order, whose indices are \(1\) and \(p\), and \(p \in M(\mathcal S)\) by hypothesis. A product of elements of a monoid lies in the monoid, so \(I(G) \subseteq M(\mathcal S)\).
Count the reciprocals. In a hypothetical partition of \(G\) into \(k \ge 2\) cosets with pairwise distinct indices \(n_i = [G:H_i]\), the block sizes sum to \(|G|\), so \(\sum_i 1/n_i = 1\). Since \(n_i \neq 1\) when \(k \ge 2\), each \(n_i\) is an element of \(I(G)\) greater than \(1\). Distinctness and the previous step give \[1 = \sum_{i=1}^{k} \frac{1}{n_i} \;\le\; \sum_{m \in I(G),\, m>1} \frac{1}{m} \;\le\; \sum_{m \in M(\mathcal S),\, m>1} \frac{1}{m} = \Lambda(\mathcal S) \;<\; 1,\] which is impossible.
Apply the criterion to the monoid, not to the group. Steps 1 and 2 say \(\mathcal J(G) \le 1 + \Lambda(\mathcal S) < 2\) for every such \(G\) at once, and the criterion then gives the conjecture for each of them.
The step that matters is uniformity. The classical submultiplicativity \(\mathcal J(G) \le \mathcal J(N)\mathcal J(G/N)\) from [1] iterates to \(\mathcal J(A_5^r) \le \mathcal J(A_5)^r = (103/60)^r\), which grows: at \(r = 2\) it is already \(10609/3600 = 2.947 > 2\), and it only gets worse. The monoid bound is one fixed number, below the threshold, for all \(r\).
Garonzi and Margolis write that their methods are insufficient for direct products of non-abelian simple groups, and they are right: their example is \(H_n = A_{p_1} \times \cdots \times A_{p_n}\), where the number of distinct simple factors grows and the generating set acquires each new small prime, so the index sum diverges. Here the factors are all copies of one fixed group \(S\). Every index of \(S^r\), for every \(r\), then lies in the single monoid \(M(\{S\})\), whose reciprocal mass is finite. Fixing the group rather than the degree is what changes the answer.
The number
The mass is not merely bounded: it is computed exactly. Every element of \(M_5\) is \(2^i3^j5^k\) with \(k \ge \kappa(i,j)\), where
\[\kappa(i,j) \;=\; \begin{cases} j - i, & j > i, \\[2pt] \max\bigl(0,\ \lceil i/2 \rceil - j\bigr), & j \le i, \end{cases}\]
so summing the geometric series in \(k\) gives \(\sum_{m \in M_5} \frac1m = \frac54 \sum_{i,j \ge 0} \frac{1}{2^i3^j5^{\kappa(i,j)}}\). The exponent lattice splits into three regions, and each sum is geometric:
| region | \(\kappa(i,j)\) | its contribution |
|---|---|---|
| \(j > i\) | \(j - i\) | \(3/35\) |
| \(\lceil i/2 \rceil \le j \le i\) | \(0\) | \(72/55\) |
| \(j < \lceil i/2 \rceil\) | \(\lceil i/2 \rceil - j\) | \(36/209\) |
Therefore
\[\sum_{m \in M_5} \frac1m \;=\; \frac54\left(\frac3{35} + \frac{72}{55} + \frac{36}{209}\right) \;=\; \frac{11463}{5852} \;=\; 1.9588174982\ldots \;<\; 2, \qquad \Lambda(\{A_5\}) = \frac{5611}{5852} = 0.9588174982\ldots\]
and the margin below the threshold is \(\frac{241}{5852} \approx 0.0412\).
An independent route confirms it without assuming the value: enumerating the monoid below \(10^{12}\) finds 1772 elements, and bounding the remainder over the superset
\[M_5 \subseteq \{2^i3^j5^k : k \ge 1,\ i \le 2j+2k\} \;\cup\; \{2^i3^j : 1 \le j \le i \le 2j\}\]
gives \(\sum_{m \in M_5} 1/m \le 1.95881749836\), which brackets the exact value from above.
A second, independent confirmation exists at \(r = 2\): the full index set \(I(A_5 \times A_5)\) is a set of 33 values, exactly the pairwise products of \(I(A_5)\), and
\[\sum_{n \in I(A_5 \times A_5),\, n>1} \frac{1}{n} = \frac{1639}{1800} < 1, \qquad \mathcal J(A_5 \times A_5) = \frac{3439}{1800} = 1.9105\ldots < 2 .\]
For \(r \ge 3\) that spectrum is out of computational reach, which is precisely why the monoid bound — valid for all \(r\) at once — is the right tool.
What else this settles
The same computation is carried out for other simple groups, with the same conclusion. Each group below has a monoid whose reciprocal mass is below 2, hence \(\Lambda < 1\), so every group whose composition factors are that group or the corresponding cyclic group is HS, direct powers included. \(A_5\) is the extreme case, with the least margin.
| Group | Order | Reciprocals over its index monoid |
|---|---|---|
| \(A_5\) | 60 | 1.95886 |
| \(\mathrm{PSL}(2,7)\) | 168 | 1.64694 |
| \(A_6\) | 360 | 1.62535 |
| \(A_7\) | 2520 | 1.45468 |
| \(A_8\) | 20160 | 1.41097 |
| \(M_{11}\) | 7920 | 1.33807 |
| \(M_{12}\) | 95040 | 1.14142 |
| \(\mathrm{Sz}(8)\) | 29120 | 1.02718 |
Mixed sets work too: \(\mathcal S = \{\mathrm{PSL}(2,7), \mathrm{PSL}(2,8)\}\) has mass \(1.87769 < 2\), so every group with composition factors in that pair is HS, including \(\mathrm{PSL}(2,7)^r\), \(\mathrm{PSL}(2,8)^s\) and all their subdirect products.
What it does not settle
- Products of different simple groups. \(A_5 \times \mathrm{PSL}(2,7)\) is not covered: a mixed finite set can have monoid mass above the threshold, and the examples in [1] show the index sums are genuinely unbounded once the factor set grows.
- Solvable groups. The conjecture is open there, and this method gives nothing: a solvable group’s composition factors are cyclic, so its monoid is generated by primes, whose reciprocal mass exceeds 1 immediately.
- Closure under products. Only one direction is known: quotients of HS groups are HS. This is a sufficient condition, not a deduction of \(A_5^r\) from \(A_5\).
- Sharpness. Whether the threshold can be pushed above 1 is open. \(\Lambda(\{A_5\}) = 0.9588\) leaves a margin of about \(0.041\), and no group in the table comes closer to the boundary.
Verification
The load-bearing inequality is checked by verify/R003.py in the private technical record: it enumerates the monoid in exact rational arithmetic, bounds the tail over the superset above, and confirms \(\sum_{m \in M_5} 1/m \le 1.95881749836 < 2\). It does not check the extension identity for indices or the criterion itself — those are proofs, not computations, and are cited above.
References
- M. Garonzi, L. Margolis, The Herzog–Schönheim conjecture for simple and symmetric groups, arXiv:2509.25118. Theorem 1.2 (symmetric groups), Theorem 1.3 (simple groups), Lemma 2.1(1) (the \(\mathcal J < 2\) criterion, attributed there to Korec and Znám, 1977), Remark 5.1 (the direct-product obstruction).
- Z.-W. Sun, On the Herzog–Schönheim conjecture for uniform covers of groups, J. Algebra 273 (2004) 153–175, arXiv:math/0306099.
- The Herzog–Schönheim conjecture for small groups and harmonic subgroups, arXiv:1803.03569 — the case \(|G| < 1440\).
- Erdős Problems #274, erdosproblems.com/274 — the conjecture listed as open.