Try
Two integer counters have formal generators e₁ and e₂. Declare that (2,0) and (2,6) may be added or removed without changing the recorded state. Which integer vectors represent the same state as zero? They are precisely integer combinations of those two relation vectors. We are not allowed arbitrary rational combinations: dividing the first relation by two would incorrectly declare e₁ zero.
Put the relations in the columns of A=[[2,2],[0,6]]. It defines a homomorphism A:Z²→Z² by matrix multiplication. Its image is the subgroup of allowed changes. The quotient Z²/AZ² identifies x and y exactly when x−y is an allowed change. Subtract the first relation from the second: the new columns are (2,0),(0,6). Predict why this operation preserves the subgroup, while dividing the first column by two does not.
Predict
In general A is an n-by-m integer matrix, a map Zᵐ→Zⁿ. The n coordinates index generators; the m columns are relations. Its cokernel, written coker A, means the abelian group Zⁿ/AZᵐ. A finite presentation specifies finitely many generators and finitely many relations of this form. It is a presentation of an abelian group, so commutativity is already part of the framework.
Use modules and maps for kernels, quotients and the Z-module interpretation. A basis here means a list with unique integer coefficients. A rational vector-space basis need not be an integer basis: (2,0),(0,1) spans Q² over Q but misses (1,0) over Z.
Predict the dimensions of the matrices multiplying A on the left and right. Left multiplication changes the ambient generator coordinates and uses an n-by-n matrix U. Right multiplication changes the relation-domain coordinates and uses an m-by-m matrix V. Their roles and sizes are different even when A happens to be square.
Experiment
Return to the two-counter example. Let V=[[1,−1],[0,1]] and U=[[1,0],[1,1]]. Both have integer inverses. Form D=AV and B=UAV, keeping the order explicit.
| Stage | Matrix, listed by rows | Meaning |
|---|
| A | (2,2); (0,6) | Original column relations |
| D=AV | (2,0); (0,6) | Replace relation 2 by relation 2 minus relation 1 |
| B=UD | (2,0); (2,6) | Send ambient coordinate (x,y) to (x,x+y) |
For D, each class has a unique representative (i,j) with 0≤i≤1 and 0≤j≤5, using integer division in each coordinate. There are twelve classes, and addition of residues identifies the quotient with Z/2Z ⊕ Z/6Z. Under U, the representative (1,5) becomes (1,6). Check that a D-relation (2a,6b) becomes the B-relation (2a,2a+6b). In contrast, dividing D’s first column by two produces diag(1,6), with only six classes. The quotient has changed.
Understand
Which matrix changes are reversible over Z?
A square integer matrix U is unimodular if it has an integer inverse. Equivalently, det U is 1 or −1. Indeed, if U and U⁻¹ have integer entries, their integer determinants multiply to 1, so each determinant is a unit of Z. Conversely, the adjugate identity gives U⁻¹=adj(U)/det U; when the determinant is ±1 all entries are integers. For the zero-dimensional square matrix, determinant 1 and the unique empty identity map give the same convention.
Swapping two rows, changing a row’s sign, or adding an integer multiple of another row is reversible by an operation of the same kind; the corresponding column operations are reversible too. Scaling a row by 2 is invertible over Q but not over Z. Some particular noninvertible operations might accidentally preserve an isomorphism type, but there is no general preservation theorem for them.
If V is unimodular, VZᵐ=Zᵐ, so AVZᵐ=AZᵐ: changing the relation-domain basis leaves exactly the same subgroup. If U is unimodular, it carries AZᵐ onto UAZᵐ. Hence
Φ: coker A → coker(UAV), [x] ↦ [Ux]
is well-defined: a difference Az becomes UAz, a relation of UAV because z=V(V⁻¹z). Conversely, if Ux−Uy=UAVz, multiplication by U⁻¹ gives x−y=AVz, an original relation. This proves injectivity and representative independence; surjectivity follows by writing any target vector as U(U⁻¹y). The inverse is [y]↦[U⁻¹y]. These are additive maps. Notice that V does not appear in the formula for the quotient isomorphism; it reorganizes relations rather than ambient coordinates.
Why finitely many relations suffice
We must prove this before claiming a classification of all finitely generated abelian groups. Every subgroup H of Zⁿ has an integer basis with at most n members. Induct on n, starting with Z⁰=0 and its empty basis. For n>0 let π be projection onto the last coordinate. The subgroup π(H) of Z is dZ for some d≥0: if nonzero, choose its least positive element d, and integer division shows every element is a multiple of d. This is the integer ideal argument from Euclidean domains.
If d=0, every vector of H has last coordinate zero. Regard H as a subgroup of Zⁿ⁻¹ and apply induction. If d>0, choose h∈H whose last coordinate is d. Let K=H∩ker π, a subgroup of Zⁿ⁻¹. Induction supplies an integer basis k₁,…,kᵣ of K, with r≤n−1. For any x∈H, its last coordinate is qd for a unique integer q. Then x−qh∈K. Therefore k₁,…,kᵣ,h spans H. If Σaᵢkᵢ+bh=0, the last coordinate gives bd=0, hence b=0, and independence in K gives every aᵢ=0. Thus this is an integer basis. This proves freeness, finite generation and the size bound together; it does not assume that H is already finitely generated.
Now let an abelian group M be generated by g₁,…,gₙ. Define the surjection φ:Zⁿ→M by φ(x)=Σxᵢgᵢ. Its kernel H is free and finitely generated by the theorem just proved. Put a basis of H in the columns of A. The map [x]↦φ(x) gives an isomorphism coker A≅M: equality of outputs is exactly difference in H, and φ is onto. If H=0 there are no columns and the cokernel is Zⁿ. If n=0 the only generated group is zero. A finite presentation need not describe a finite group: Z itself is a basic counterexample.
Apply
Change the relation matrix to A′ with rows (3,3),(0,9),(0,0). There are three generators but two relations. Subtract column 1 from column 2, then interchange ambient coordinates 1 and 3. Compute the new matrix and the quotient isomorphism. Which coordinate remains unrestricted?
Check the rectangular presentation
The column subtraction gives D′ with rows (3,0),(0,9),(0,0). The row interchange gives B′ with rows (0,0),(0,9),(3,0). The quotient isomorphism sends [(x,y,z)] to [(z,y,x)]. In D′ coordinates the first coordinate is taken modulo 3, the second modulo 9 and the third is an arbitrary integer. Thus coker A′≅Z/3Z ⊕ Z/9Z ⊕ Z. In B′ coordinates the free coordinate has moved to the first position. The group is infinite despite having a finite presentation; its torsion subgroup has 27 elements. Dividing the relation column (3,0,0) by 3 would remove the order-three factor and is not a reversible integer operation.
For a separate proof action, apply the subgroup induction to H generated by (2,0),(1,3). Its last-coordinate image is 3Z; choose h=(1,3). Show directly that K is generated by (2,0), and recover each vector as its unique K component plus a multiple of h. Explain why the proof generalizes even when no initial finite generating list for H is given.
Continue
Continue to Smith reduction, which will construct reversible operations leading to a divisibility-ordered diagonal. This lesson proves what operations preserve the quotient and why finite relation matrices exist; it does not yet prove that the reduction terminates or that its diagonal factors are unique.
The existing finite abelian structure theorem provides a complementary structural proof. Integer presentations add a constructive way to begin with generators and relations, including groups with a free part. Keep the matrix identity, the induced quotient map and the classification conclusion as separate claims that each need evidence.
