False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: nonisomorphic finite groups always have different character tables

Statement

The statement "nonisomorphic finite groups always have different character tables" is false: the quaternion group $Q_{8}$ and the dihedral group $Dih (C_{4})$ are nonisomorphic finite groups with the same character table.

Facts & Assumptions

Given: The groups $Q_{8}$ and $Dih (C_{4})$ , both of order $8$ .

[F1]

The character table of $Q_{8}$ has five rows with values $(1, 1, 1, 1, 1)$ , $(1, 1, 1, - 1, - 1)$ , $(1, 1, - 1, 1, - 1)$ , $(1, 1, - 1, - 1, 1)$ , and $(2, - 2, 0, 0, 0)$ on the classes of $1$ , $- 1$ , $i$ , $j$ , $k$ (The character table of $Q_{8}$ ).

[F2]

The character table of $Dih (C_{4})$ has five rows with values $(1, 1, 1, 1, 1)$ , $(1, 1, - 1, 1, - 1)$ , $(1, 1, 1, - 1, - 1)$ , $(1, 1, - 1, - 1, 1)$ , and $(2, - 2, 0, 0, 0)$ on the classes of $1$ , $r^{2}$ , $r$ , $s$ , $s r$ (The character table of $Dih (C_{4})$ ).

[F3]

$- 1$ is the only element of order $2$ in $Q_{8}$ ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[F4]

In $Dih (C_{4})$ , the elements $r^{2}$ and $s$ both have order $2$ ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

Refutation

technique · construct

1.1

The row sets of [F1] and [F2] are identical as multisets of tuples, so after a row permutation the two tables agree entry for entry.

F1F2given

1.2

By [F3], $Q_{8}$ has exactly one element of order $2$ , while by [F4] $Dih (C_{4})$ has at least the two elements $r^{2}$ and $s$ of order $2$ . The number of elements of order $2$ is invariant under isomorphism, so the two groups are not isomorphic.

F3F4given

2.1

The two tables have the same number of rows and columns with the same class sizes $1$ , $1$ , $2$ , $2$ , $2$ , so they are the same character table up to the labelling of the groups.

step 1.1algebra

3.1

Steps 2.1 and 1.2 exhibit nonisomorphic finite groups with the same character table, so the claimed statement is refuted.

step 2.1step 1.2discharge-construct: counterexample∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Webb, A Course in Finite Group Representation Theory, Chapter 3 (standard reference, not scraped)