# Finitely-Generated Free Abelian Groups

An Interactive Primer on Homomomorphisms between Free Abelian Groups

Algebraic Topology
Interactive Tutorial
Author

Sushovan Majhi

Published

July 26, 2021

Let $$(G,+)$$ and $$(G',+)$$ be finitely-generated free (additive) abelian groups of dimension $$m$$ and $$n$$, respectively. We choose arbitrary ordered bases $$\mathcal{E}=(e_1,e_2,\ldots,e_m)$$ and $$\mathcal{E}'=(e_1',e_2',\ldots,e_n')$$ for $$G$$ and $$G'$$, respectively.

## Matrix of A Homomorphism

A homomorphism $$F:G\to G'$$ is a linear function, i.e., $$F(x+y)=F(x)+F(y)$$ for all $$x,y\in G$$.

As a consequence, $$F$$ is uniquely determined by its values on the elements of $$\mathcal{E}$$. For each $$1\leq j\leq m$$, the image $$F(e_j)$$ can be written uniquely as a (formal) linear combination of the elements of $$\mathcal{E}'$$. Let $F(e_j) = \sum\limits_{i=1}^{n} a_{ij}e_i'.$

The matrix $$A:=\{a_{ij}\}\in\mathcal{M}_{n,m}(\mathbb Z)$$ is called the matrix of $$F$$ w.r.t. the (ordered) bases $$\mathcal{E},\mathcal{E}'$$, and is denoted by $$[F]_{\mathcal{E},\mathcal{E'}}$$. In matrix notation, we have $\bigg[F(e_1),\ldots,F(e_j),\ldots,F(e_m)\bigg]= \bigg[e_1',\ldots,e_i',\ldots,e_n'\bigg] \begin{bmatrix} a_{11} & \ldots & a_{1j} & \ldots & a_{1m} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{i1} & \ldots & a_{ij} & \ldots & a_{im} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{n1} & \ldots & a_{nj} & \ldots & a_{nm} \\ \end{bmatrix}.$ Tolerating a mild abuse of notation, we can also express the above relation as $$F(\mathcal{E})=\mathcal{E'}[F]_{\mathcal{E},\mathcal{E'}}$$.

### Example:

Let $$F:\mathbb{Z}^3\to \mathbb{Z}^2$$ be a homomorphism defined by $F(a,b,c)=(3a+4b-2c,2b+5c).$

A natural choice of ordered bases is $$\mathcal{E}=\bigg((1,0,0),(0,1,0),(0,0,1)\bigg)$$ and $$\mathcal{E}'=\bigg((1,0),(0,1)\bigg)$$. In order to compute $$[F]_{\mathcal{E},\mathcal{E}''}$$, the matrix of $$F$$ w.r.t. the chosen bases, we note that \begin{aligned} F(1,0,0) &= (3,0) = 3(1,0) + 0(0,1) \\ F(0,1,0) &= (4,2) = 4(1,0) + 2(0,1) \\ F(0,0,1) &= (-2,5) = -2(1,0) + 5(0,1) \end{aligned} So, the matrix of $$F$$ is $\begin{bmatrix} 3 & 4 & -2\\ 0 & 2 & 5 \end{bmatrix}.$

Question What would the matrix of $$F$$ be if we chose the bases $$\bigg((1,1,0),(1,0,1),(0,1,1)\bigg)$$ and $$\bigg((1,0),(1,1)\bigg)$$?

Following the technique (basically the definition) of the above example we can compute the matrix of $$F$$ under the new set of bases.

## Change of Basis

Let $$\mathcal{B}=(e_1,e_2,\ldots,e_m)$$ be an ordered basis of a finitely-generated free abelian group $$G$$. Now we consider an new basis $$\mathcal{C}=(f_1,f_2,\ldots,f_m)$$ of $$G$$. The basechange matrix encodes the change the basis from (old) $$\mathcal{B}$$ to (new) $$\mathcal{C}$$. The basechange matrix facilitates the computation of the matrix of a homomomorphism w.r.t. a set of new bases.

The starting point is to write the elements of the new basis as a linear combination of those of the old basis. For $$j=1,2,\ldots,m'$$, $f_j = \sum\limits_{i=1}^{m} p_{ij}e_i.$

The matrix $$P$$ is called the basechange matrix the order bases $$\mathcal{B}\text{ and }\mathcal{C}$$. In matrix notation, $\mathcal{C}=\mathcal{B}P.$

As we can immediately see, the basechange matrix is unique, and it is also invertible [artin], i.e., $$P\in GL_{m}(\mathbb Z)$$. Given an invertible, integer matrix, one can also use the above matrix relation to compute the new basis.

### Example:

For this example, I let you interact with the tutorial. We choose the dimension of $$G$$ using the following slider. Upon choosing a value, you readily see the initial basis:

and randomly generated basechange matrix $$P$$.

Now, we take a random invertible matrix of the chosen dimension:

The new basis of $$G$$ is:

### Application to Homomorphisms

Let $$F:G\to G'$$ be a homomorphism. Let $$A$$ denote the matrix of $$F$$ w.r.t. a pair of old ordered bases $$(\mathcal{B},\mathcal{B}')$$ for $$G$$ and $$G'$$, respectively. If a new pair of order bases $$(\mathcal{C},\mathcal{C}')$$ chosen, one would be interested to compute the matrix of $$F$$ w.r.t. the new bases.

Let $$P\in\mathcal{M}_m(\mathbb Z)$$ and $$Q\in\mathcal{M}_n(\mathbb Z)$$ denote the basechange matrices for $$G$$ and $$G'$$, respectively. So, $\mathcal{C}=\mathcal{B}P\text{, and }\mathcal{C}'=\mathcal{B}'Q.$

By the definition of $$A$$, note that $$F(\mathcal{B})=\mathcal{B}'A$$. Therefore, \begin{aligned} F(\mathcal{B})P &=(\mathcal{B}'A)P \\ &= (\mathcal{C}'Q^{-1})AP \\ &= \mathcal{C}'(Q^{-1}AP) \end{aligned}

Using the fact that $$F$$ is a homomomorphism and letting $$\mathcal{B}=(e_1,e_2,\ldots,e_m)$$ and $$\mathcal{C}=(f_1,f_2,\ldots,f_m)$$, we simplify the LHS. \begin{aligned} F(\mathcal{B})P &= \big[F(e_1),\ldots,F(e_j),\ldots,F(e_m)\big]P \\ &= \left[\sum_{j=1}^mF(e_j)p_{1j},\ldots,\sum_{j=1}^mF(e_j)p_{mj}\right] \\ &= \left[F\left(\sum_{j=1}^m p_{1j}e_j\right),\ldots,F\left(\sum_{j=1}^m p_{1j}e_j\right)\right] \\ &= \left[F\left(f_1\right),\ldots,F\left(f_m\right)\right] \\ &= F(\mathcal{C}). \end{aligned}

Therefore, $F(\mathcal{C})=\mathcal{C}'\left(Q^{-1}AP\right).$

We conclude that $$\left(Q^{-1}AP\right)$$ is the matrix of $$F$$ w.r.t. the new pair of basis $$(\mathcal{C},\mathcal{C}')$$.