Number of groups with a given spectrum

DefinitionThe spectrum of a group (Kourovka 12.84)#✓fully proved

The spectrum of a group \(G\) is the set \(\{\operatorname{ord}(g) : g \in G\}\) of the orders of its elements, where an element of infinite order is counted as having order \(0\).

def elementOrders (G : Type*) [Group G] : Set ℕ :=
  Set.range (orderOf : G → ℕ)

Used by 1

Used bySyskin's question on groups with a given spectrum (Kourovka 12.84)

DefinitionSyskin's question on groups with a given spectrum (Kourovka 12.84)#✓fully proved

The proposition that for every set \(\Omega\) of natural numbers, if two non-isomorphic finite groups both have set of element orders \(\Omega\), then for every natural number \(k\) there are \(k\) pairwise non-isomorphic finite groups whose set of element orders is \(\Omega\); that is, infinitely many.

def twoRealizationsImplyInfinitelyMany : Prop :=
  ∀ Ω : Set ℕ,
    (∃ (G H : Type) (_ : Group G) (_ : Finite G) (_ : Group H) (_ : Finite H),
      elementOrders G = Ω ∧ elementOrders H = Ω ∧ IsEmpty (G ≃* H)) →
    ∀ k : ℕ, ∃ (G : Fin k → Type) (_ : ∀ i, Group (G i)) (_ : ∀ i, Finite (G i)),
      (∀ i, elementOrders (G i) = Ω) ∧ ∀ i j, i ≠ j → IsEmpty (G i ≃* G j)

Uses 1 · Used by 1

Statement usesThe spectrum of a group (Kourovka 12.84)

Used byTwo isospectral finite groups do not force infinitely many

TheoremTwo isospectral finite groups do not force infinitely many#●ready to prove

There are a set \(\Omega\) of natural numbers and a natural number \(k\) such that two non-isomorphic finite groups have set of element orders \(\Omega\), but there are no \(k\) pairwise non-isomorphic finite groups whose set of element orders is \(\Omega\).

For example, by Theorem 2.1(iii) of the survey of Grechkoseeva, Mazurov, Shi, Vasil'ev and Yang, the finite groups with the set of element orders of \(S_6(2)\) are exactly \(S_6(2)\) and \(O_8^+(2)\), which have different orders. The Kourovka Archive answers 12.84 "No, it is not" (V. D. Mazurov, 1994).

public theorem_wanted twoRealizationsImplyInfinitelyMany_refuted :
    ¬ twoRealizationsImplyInfinitelyMany

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 12.84 (well-known problem; S. A. Syskin); resolution: V. D. Mazurov, Algebra and Logic 33 (1994), no. 1, 49-55; M. A. Grechkoseeva, V. D. Mazurov, W. Shi, A. V. Vasil'ev and N. Yang, Finite groups isospectral to simple groups, arXiv:2111.15198, Kourovka Archive, Problem 12.84: No, it is not; Grechkoseeva et al., Theorem 2.1(iii): a finite group has the spectrum of S_6(2) if and only if it is S_6(2) or O_8^+(2), https://arxiv.org/abs/2111.15198.

Uses 1

Statement usesSyskin's question on groups with a given spectrum (Kourovka 12.84)

DefinitionSpectrum of a group (Kourovka 13.63)#✓fully proved

For a group \(G\), the spectrum \(\pi_e(G)\) is the set of orders of elements of \(G\), where an element of infinite order counts as having order \(0\).

def elementOrders (G : Type*) [Group G] : Set ℕ :=
  Set.range (orderOf : G → ℕ)

Used by 2

Used byAt most \(k\) isomorphism types with spectrum \(\Gamma\) · Infinitely many isomorphism types with spectrum \(\Gamma\)

DefinitionAt most \(k\) isomorphism types with spectrum \(\Gamma\)#✓fully proved

For a set \(\Gamma\) of natural numbers and a natural number \(k\), this property holds when, for any finite groups \(G_0, \dots, G_k\) (not necessarily distinct) whose spectra \(\pi_e(G_i)\) (the sets of orders of elements of \(G_i\)) all equal \(\Gamma\), there are indices \(i \ne j\) with \(G_i \cong G_j\).

This expresses \(h(\Gamma) \le k\), where \(h(\Gamma)\) is the number of isomorphism classes of finite groups with spectrum \(\Gamma\), without counting isomorphism classes directly.

def AtMostIsoClasses (Γ : Set ℕ) (k : ℕ) : Prop :=
  ∀ (G : Fin (k + 1) → Type) [∀ i, Group (G i)] [∀ i, Finite (G i)],
    (∀ i, elementOrders (G i) = Γ) → ∃ i j, i ≠ j ∧ Nonempty (G i ≃* G j)

Uses 1 · Used by 1

Statement usesSpectrum of a group (Kourovka 13.63)

Used byKourovka Problem 13.63 (bounded or infinite number of groups per spectrum)

DefinitionInfinitely many isomorphism types with spectrum \(\Gamma\)#✓fully proved

For a set \(\Gamma\) of natural numbers, this property holds when, for every natural number \(n\), there are \(n\) pairwise non-isomorphic finite groups \(G_0, \dots, G_{n-1}\) whose spectra \(\pi_e(G_i)\) (the sets of orders of elements of \(G_i\)) all equal \(\Gamma\).

This expresses \(h(\Gamma) = \infty\), where \(h(\Gamma)\) is the number of isomorphism classes of finite groups with spectrum \(\Gamma\).

def InfinitelyManyIsoClasses (Γ : Set ℕ) : Prop :=
  ∀ n : ℕ, ∃ (G : Fin n → Type) (_ : ∀ i, Group (G i)) (_ : ∀ i, Finite (G i)),
    (∀ i, elementOrders (G i) = Γ) ∧ ∀ i j, i ≠ j → IsEmpty (G i ≃* G j)

Uses 1 · Used by 1

Statement usesSpectrum of a group (Kourovka 13.63)

Used byKourovka Problem 13.63 (bounded or infinite number of groups per spectrum)

DefinitionKourovka Problem 13.63 (bounded or infinite number of groups per spectrum)#✓fully proved

The proposition, asked about by W. J. Shi, that there is a natural number \(k\) such that for every set \(\Gamma\) of natural numbers, either \(h(\Gamma) \le k\) or \(h(\Gamma) = \infty\), where \(h(\Gamma)\) is the number of isomorphism classes of finite groups \(G\) whose spectrum \(\pi_e(G)\) (set of element orders) is \(\Gamma\). Here \(h(\Gamma) \le k\) means that among any \(k + 1\) finite groups with spectrum \(\Gamma\) two are isomorphic, and \(h(\Gamma) = \infty\) means that for every \(n\) there are \(n\) pairwise non-isomorphic finite groups with spectrum \(\Gamma\).

def spectrumGroupCountBoundedOrInfinite : Prop :=
  ∃ k : ℕ, ∀ Γ : Set ℕ, AtMostIsoClasses Γ k ∨ InfinitelyManyIsoClasses Γ

Uses 2 · Used by 1

Statement usesAt most \(k\) isomorphism types with spectrum \(\Gamma\) · Infinitely many isomorphism types with spectrum \(\Gamma\)

Used byNo uniform bound on the number of groups with a given spectrum

TheoremNo uniform bound on the number of groups with a given spectrum#●ready to prove

For a set \(\Gamma\) of natural numbers, let \(h(\Gamma)\) be the number of isomorphism classes of finite groups \(G\) whose spectrum \(\pi_e(G)\) (set of element orders) is \(\Gamma\); \(h(\Gamma) \le k\) means that among any \(k + 1\) finite groups with spectrum \(\Gamma\) two are isomorphic, and \(h(\Gamma) = \infty\) means that for every \(n\) there are \(n\) pairwise non-isomorphic finite groups with spectrum \(\Gamma\). Kourovka Problem 13.63 has a negative answer: there is no natural number \(k\) such that for every set \(\Gamma\) of natural numbers, either \(h(\Gamma) \le k\) or \(h(\Gamma) = \infty\).

Zavarnitsine proves \(h(\pi_e(L_3(7^{3^r}))) = r + 1\) for every \(r \ge 0\), where \(L_3(q)\) is the projective special linear group \(\mathrm{PSL}_3(q)\), so \(h\) takes every positive finite value and no \(k\) works.

public theorem_wanted spectrumGroupCountBoundedOrInfinite_refuted :
    ¬ spectrumGroupCountBoundedOrInfinite

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 13.63 (W. J. Shi); resolution: A. V. Zavarnitsine, J. Group Theory 7 (2004), no. 1, 81-97, h(π_e(L_3(7^(3^r)))) = r + 1 for every r ≥ 0; Kourovka Archive, Problem 13.63 (answer: No, there is no such number).

Uses 1

Statement usesKourovka Problem 13.63 (bounded or infinite number of groups per spectrum)

6 fully proved2 ready to prove

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