The Linear Algebra of Free Abelian Groups Recall that a non-trivial free abelian group with basis X = {x1, x2, ··· , xd} is iso- morphic to Z × Z × · · · × Z, where we have d factors of Z. To simplify notation, we will consider the free abelian groups Z × Z × Z × Z and Z × Z × Z with bases X = {x1, x2, x3, x4} and Y = {y1, y2, y3}, respectively. However, the results in this note are valid in general. A function φ : Z × Z × Z × Z → Z × Z × Z will be a homomorphism exactly if φ(α1x1 + α2x2 + α3x3 + α4x4) = α1φ(x1) + α2φ(x2) + α3φ(x3) + α4φ(x4) for all αi ∈ Z. That is, φ will be a homomorphism exactly if it is linear with respect to integer coefficients. Since φ is determined by its values on the generators of its domain, that is, by the values φ(xi), we only have to express each such φ(xi) in terms of the basis Y , in order to fully describe φ. Since Y = {y1, y2, y3} is a basis for Z × Z × Z, there are unique integers aji such that φ(x1) = a11y1 + a21y2 + a31y3 φ(x2) = a12y1 + a22y2 + a32y3 φ(x3) = a13y1 + a23y2 + a33y3 (1) φ(x4) = a14y1 + a24y2 + a34y3 We assemble the transpose of these entries into a matrix a11 a12 a13 a14 A = a21 a22 a23 a24 . a31 a32 a33 a34 Proposition 1. Let g ∈ Z × Z × Z × Z. Then φ(g) can be computed as follows. Express g = α1x1 + α2x2 + α3x3 + α4x4 in the basis X and compute α1 a11 a12 a13 a14 β1 α2 a21 a22 a23 a24 = β2 . α3 a31 a32 a33 a34 β3 α4 Then φ(g) = β1y1 + β2y2 + β3y3 in the basis Y . Proof. 4 ! 4 4 3 X X X X φ(g) = φ αixi = αiφ(xi) = αi ajiyj i=1 i=1 i=1 j=1 3 4 ! 3 X X X = ajiαi yj = βjyj j=1 i=1 j=1 1 Proposition 2. Changing either of the bases X = {x1, x2, x3, x4} or Y = {y1, y2, y3} has the following effect on the matrix A. For simplicity, we choose specific locations for the changes to occur. However, the results clearly remain valid in general. • Replacing X by X = {x1, x2, −x3, x4} changes the entries of the third column of A to their additive inverse. • Replacing Y by Y = {y1, −y2, y3} changes the entries of the second row of A to their additive inverse. • Replacing X by X = {x3, x2, x1, x4} interchanges column one and column three. • Replacing Y by Y = {y3, y2, y1} interchanges row one and three. • Replacing X by X = {x1, x2, x3 + 5x1, x4} adds 5 times column one to column three. • Caution: replacing Y by Y = {y1 − 5y3, y2, y3} adds 5 times row one to row three. Proof. We only prove the last two statements. The remaining four statements are left as an exercise. Suppose we replace X by X = {x1, x2, x3 + 5x1, x4}. Then Equation (1) turns into φ(x1) = a11y1 + a21y2 + a31y3 φ(x2) = a12y1 + a22y2 + a32y3 φ(x3 + 5x1) = (a13 + 5a11)y1 + (a23 + 5a21)y2 + (a33 + 5a31)y3 φ(x4) = a14y1 + a24y2 + a34y3 So, the entries of A change as predicted, since A is the transpose of these entries. However, if replace Y by Y = {y1 − 5y3, y2, y3}, we get φ(x1) = a11(y1 − 5y3) + a21y2 + (a31 + 5a11)y3 φ(x2) = a12(y1 − 5y3) + a22y2 + (a32 + 5a12)y3 φ(x3) = a13(y1 − 5y3) + a23y2 + (a33 + 5a13)y3 φ(x4) = a14(y1 − 5y3) + a24y2 + (a34 + 5a14)y3, which amounts to adding five times row one to row three in the matrix A. Remark. There is no change of bases that would result in a row or a column to be multiplied by an integer other than ±1. 2 Proposition 3. Using the above changes of bases, the matrix A can be brought into the following form: n1 0 0 0 0 n2 0 0 , 0 0 n3 0 with ni > 0 and n1|n2|n3. Proof. We may assume that there is an entry different from zero. Let a be a non- zero entry of the matrix. By interchanging rows and columns we can move a into the upper left corner of the matrix. Make a positive by changing the entire first row to their additive inverse, if necessary. Now, if b is any entry in the first column, below a, we can divide b = qa + r with 0 6 r < a. By adding (−q) times the first row to the row containing b, we can therefore make r appear in place of b. If r is not zero, then it is smaller than a. If that happens, we switch rows, so as to place r into the upper left corner instead of a. Every time we do this, the entry in the upper left corner decreases. So, this cannot go on forever. We must reach a point where the remainder r is zero for all the entries in the first row below the upper left corner. So, we can now assume that we have zeroes in the entire first column below a. Arguing in the other direction, we can arrange for the first row to consist of zeroes, to the right of a. We now arrange for a to divide all the remaining entries in the matrix. Let c be any entry of the matrix, other than a. We divide with remainder: c = qa + r where 0 6 r < a. Then we apply some clever row and column operations to replace c by r. (Which ones?) If r is not zero, we can move it into the upper left corner of the matrix where it will replace a. Again, we clear all entries below and to the right. This procedure keeps reducing the size of our upper left entry. So, sooner or later, we can only get remainders of zero. We have now found n1, namely n1 = a, because we have reached the point, where the matrix looks like this n1 0 0 0 0 0 0 0 a22 a23 a24 0 0 0 0 a32 a33 a34 and n1 divides all other entries. To finish the proof, simply apply the same logic to the smaller matrix 0 0 0 a22 a23 a24 0 0 0 . a32 a33 a34 3 Proposition 4. Suppose there are bases with respect to which the homomorphism φ : Z × Z × Z × Z → Z × Z × Z has a matrix of the form n1 0 0 0 0 n2 0 0 , 0 0 n3 0 with ni > 0 (and n1|n2|n3). Then Z × Z × Z/Im φ is isomorphic to Zn1 × Zn2 × Zn3 . Proof. Let X = {x1, x2, x3, x4} and Y = {y1, y2, y3} be bases for Z × Z × Z × Z and Z × Z × Z, respectively, with respect to which φ has the given matrix. This means that φ(x1) = n1y1, φ(x2) = n2y2, φ(x3) = n3y3 and φ(x4) = 0. Since the basis X = {x1, x2, x3, x4} generates the domain of φ, we see that {φ(x1), φ(x2), φ(x3)} generates its image H = Im φ. That is, H = h{n1y1, n2y2, n3y3}i. Since Z × Z × Z is abelian, H is a normal subgroup of Z × Z × Z. Consider the homomorphism f : Z × Z × Z → Z × Z × Z/H given by f(α1, α2, α3) = α1y1 + α2y2 + α3y3 + H. Then f is onto, because Y = {y1, y2, y3} is a basis for Z × Z × Z. We claim that the kernel of f is generated by the elements (n1, 0, 0), (0, n2, 0), (0, 0, n3), which will finish the proof by the First Isomorphism Theorem, because then ∼ ∼ Zn1 × Zn2 × Zn3 = Z × Z × Z/ h{(n1, 0, 0), (0, n2, 0), (0, 0, n3)}i = Z × Z × Z/H. Observe that each of these elements is in the kernel of f. For example, f(n1, 0, 0) = n1y1 + H = 0 + H, since n1y1 ∈ H. Hence every combination of the elements (n1, 0, 0), (0, n2, 0), (0, 0, n3) is in the kernel of f. Conversely, if f(α1, α2, α3) = 0 + H, then α1y1 + α2y2 + α3y3 + H = 0 + H. So, α1y1 + α2y2 + α3y3 ∈ H, which implies that α1y1 + α2y2 + α3y3 = γ1n1y1 + γ2n2y2 + γ3n3y3 for some γi ∈ Z. However, since Y = {y1, y2, y3} is a basis, this implies that αi = γini for i = 1, 2, 3. Hence (α1, α2, α3) = γ1(n1, 0, 0) + γ2(0, n2, 0) + γ3(0, 0, n3). Remark. Since 1 divides every integer and every integer divides 0, it might happen that some of the initial ni’s are 1 and that some of the ni’s at the end are 0. Note ∼ that Z1 = {0}, whereas Z0 = Z. Remark. Other scenarios are handled similarly. Suppose, the matrix for a homo- morphism φ : Z × Z × Z → Z × Z × Z × Z reduces to n1 0 0 0 n2 0 , 0 0 n3 0 0 0 ∼ with ni > 0 (and n1|n2|n3). Then Z×Z×Z×Z/Im φ = Zn1 ×Zn2 ×Zn3 ×Z, because X = {x1, x2, x3}, Y = {y1, y2, y3, y4}, and Im φ = h{n1y1, n2y2, n3y3, 0y4}i. 4 Example. Classify Z3 × Z × Z6/ h{([3], 4, [2]), ([2], 0, [5])}i in invariant factor form. ∼ Step 1: G = Z × Z × Z/H with H = h{(3, 0, 0), (0, 0, 6), (3, 4, 2), (2, 0, 5)}i. Step 2: Consider the homomorphism φ : Z × Z × Z × Z → Z × Z × Z given by φ(α1, α2, α3, α4) = α1(3, 0, 0) + α2(0, 0, 6) + α3(3, 4, 2) + α4(2, 0, 5).