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ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-11
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Amalgamating infinite cyclic groups by multiplication by m and n gives the presentation with relation x^m=y^n

Example

For positive integers m,n, amalgamating infinite cyclic groups ⟨x⟩ and ⟨y⟩ along maps sending a generator to xm and yn gives ⟨x,y∣xm=yn⟩. Both cyclic factor maps remain injective.

Facts & Assumptions

Given: The objects and hypotheses in the example.

[L1]

Let G=⟨X∣R⟩ and H=⟨Y∣S⟩ with disjoint generators, and let f,h embed K. If T generates K and words ut(X),vt(Y) represent f(t),h(t), then G∗KH≅⟨X⊔Y∣R∪S∪{utvt−1:t∈T}⟩. (A free product with amalgamation has the factor presentations plus the amalgamating relations).

[L2]

The canonical maps G→G∗KH and H→G∗KH are injective. (The factor maps into a free product with amalgamation are injective).

[L3]

Every class in W(X)/∼ contains exactly one reduced word. (Every class in W(X)/∼ contains exactly one reduced word).

[L4]

For every set X, the group Fword(X)=W(X)/∼ together with iword(x)=[x] is a free group on X in the sense of def-free-group. (The word-quotient group W(X)/∼ satisfies the universal property of the free group on X).

[L5]

Let F(X) be a free group and let R⊆F(X) be a set of words, called relations. The group with presentation ⟨X∣R⟩:=F(X)/⟨ ⁣⟨R⟩ ⁣⟩F(X) is the quotient by the normal closure of R. The members of X are its generators. In this quotient, every relation in R becomes the identity, as do all consequences forced by normality. (Group presentation by generators and relations).

[L6]

Let G be a group and R⊆G. Then ⟨ ⁣⟨R⟩ ⁣⟩G={g1r1ε1g1−1⋯gnrnεngn−1:n∈N, gi∈G, ri∈R, εi∈{1,−1}}. For n=0 the displayed product is the identity. Replacing every conjugator gi by gi−1 gives the equivalent convention gi−1riεigi. (The normal closure of R is the set of finite products of conjugates of elements of R and their inverses).

Verification

technique · direct
1.1

The one-generator empty-relator word model is infinite cyclic: its reduced words are the distinct powers of its generator, and the word-quotient freeness theorem gives the singleton universal property.

givenL1L2L3L4L5L6
2.1

The maps from the edge group are injective because m,n>0 and the factors have infinite order.

step 1.1
3.1

The amalgamated-presentation theorem adds exactly the relation xmy−n=e, equivalently xm=yn, and the factor-embedding theorem preserves both cyclic factors.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources