Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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FALSE: canonical factor maps into every group pushout are injective

Statement

False claim: both canonical factor maps into every group pushout are injective.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

For a pushout of f:KGf:K\to G and h:KHh:K\to H, iH(h(kerf))={e},iG(f(kerh))={e}.i_H(h(\ker f))=\{e\},\qquad i_G(f(\ker h))=\{e\}. Hence canonical factor maps in an arbitrary group pushout need not be injective. No equality with their full kernels is asserted. (The kernels of the amalgamating maps are killed in the opposite canonical maps to a group pushout).

[L2]

For homomorphisms f:KGf:K\to G and h:KHh:K\to H, let NN be the normal closure in GHG\ast H of {jG(f(k))jH(h(k))1:kK}.\{j_G(f(k))j_H(h(k))^{-1}:k\in K\}. Then (GH)/N(G\ast H)/N, with the induced factor maps jGj_G and jHj_H, is a pushout of ff and hh. (A group pushout is the quotient of a free product by the amalgamating relations).

[L3]

For every nNn\in\mathbb N, view nn as its canonical nonnegative integer and put nZ:={nk:kZ}n\mathbb Z:=\{nk:k\in\mathbb Z\}. Then the left cosets of nZn\mathbb Z in (Z,+)(\mathbb Z,+) are exactly the congruence classes modulo nn, and coset addition is the published addition of congruence classes. Thus (Z,+)/nZ=(Z/n,+)(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+) as the same group on the same underlying set. This includes n=0n=0 and n=1n=1. (For every nNn\in\mathbb N, the congruence-class group (Z/n,+)(\mathbb Z/n,+) is the quotient group (Z,+)/nZ(\mathbb Z,+)/n\mathbb Z).

Refutation

technique · direct
1.1

Take K=C2K=C_2, G=1G=1, H=C2H=C_2, with f:KGf:K\to G trivial and h:KHh:K\to H the identity.

givenL1L2L3
2.1

The amalgamating relation kills the generator of HH, so the quotient construction makes the pushout trivial. Equivalently, kernel collapse kills h(kerf)=Hh(\ker f)=H.

step 1.1
3.1

The canonical map H=C21H=C_2\to1 is not injective, refuting the claim.

step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 61 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources