Induced_Group_Product_is_Homomorphism_iff_Commutative
Disproved⚠️ Retired — specification defect
The Lean statement below does not encode the problem shown on this page, so its
Disprovedstatus carries no information about that problem. Do not import this node or use it as a dependency.
If elements from two subgroups commute with each other, then the product map is a homomorphism.
Why this node was retired
The posted statement is
theorem Induced_Group_Product_is_Homomorphism_iff_Commutative {G : Type _} [Group G] (a b c d : G) (hab : a * c = c * a) (hbd : b * d = d * b) : (a * b) * (c * d) = (a * c) * (b * d) := by sorry
The hypotheses constrain the wrong pairs; the middle factors b,c must commute.
The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.
Proposed corrected statement
For a,b,c,d in a group, bc=cb implies (ab)(cd)=(ac)(bd). The group-level multiplication H×K→G is a homomorphism exactly when every h∈H commutes with every k∈K.
Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.
import Mathlib.Algebra.Group.Defs
theorem Induced_Group_Product_is_Homomorphism_iff_Commutative {G : Type _} [Group G] (a b c d : G) (hab : a * c = c * a) (hbd : b * d = d * b) : (a * b) * (c * d) = (a * c) * (b * d) := by sorry