Homology: Free Abelian Groups

 Before we study homology groups, we review some basics of abelian group theory in this note.

The group operation for an abelian group is denoted by $+$. The unit element is denoted by $0$

Let $G_1$ and $G_2$ be abalian groups. A map $f: G_1\longrightarrow G_2$ is said to be a homomorphism if $$f(x+y)=f(x)+f(y),\ x,y\in G_1.$$ If $f$ is also a bijection (i.e one-to-one and onto), $f$ is called an isomorphism. If there is an isomorphism $f: G_1\longrightarrow G_2$, $G_1$ is said to be isomorphic to $G_2$ and we write $G_1\stackrel{f}{\cong} G_2$ or simply $G_1\cong G_2$.

Example. Define a map $f: \mathbb{Z}\longrightarrow\mathbb{Z}_2=\{0,1\}$ by $$f(2n)=0\ \mbox{and}\ f(2n+1)=1.$$ Then $f$ is a homomorphism.

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$. If $H$ is a subgroup of $G$, we write $H\leq G$.

Example. For any $k\in\mathbb N$, $k\mathbb{Z}=\{kn: n\in\mathbb{Z}\}\leq\mathbb{Z}$.

Example. $\mathbb{Z}_2=\{0,1\}\not\leq\mathbb{Z}$.

Let $H\leq G$. Define a relation on $G$ by
$$\forall x,y\in G,\ x\sim y\ \mbox{if}\ x-y\in H.$$
Then $\sim$ is an equivalence relation on $G$. The equivalence class of $x\in G$ is denoted by $[x]$, i.e. \begin{align*}
[x]&=\{y\in G: y\sim x\}\\
&=\{y\in G: y-x\in H\}.
\end{align*}
Let $G/H$ be the quotient set
$$G/H=\{[x]: x\in G\}.$$
Define an operation $+$ on $G/H$ by
$$[x]+[y]=[x+y],\ \forall [x],[y]\in G/H.$$
Then $G/H$ becomes an abelian group with this operation.

Example. $\mathbb{Z}/2\mathbb{Z}=\{[0],[1]\}$. Define $\varphi: \mathbb{Z}/2\mathbb{Z}\longrightarrow\mathbb{Z}_2$ by
$$\varphi([0])=0\ \mbox{and}\ \varphi([1])=1.$$
Then $\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}_2$. In general, for every $k\in\mathbb N$, $\mathbb{Z}/k\mathbb{Z}\cong\mathbb{Z}_k$.

Lemma. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then

1. $\ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0)$ is a subgroup of $G_1$.
2. $\mathrm{im}f=\{f(x): x\in G_1\}$ is a subgroup of $G_2$.
   

Theorem. [Fundamental Theorem of Homomorphism]
Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then
$$G_1/\ker f\cong\mathrm{im}f.$$

Example. Let $f: \mathbb{Z}\longrightarrow\mathbb{Z}_2$ be defined by
$$f(2n)=0,\ f(2n+1)=1.$$
Then $\ker f=2\mathbb{Z}$ and $\mathrm{im}f=\mathbb{Z}_2$. By Fundamental Theorem of Homomorphism,
$$\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}_2.$$

Take $r$ elements $x_1,x_2,\cdots,x_r$ of $G$. The elements of $G$ of the form
$$n_1x_1+n_2x_2+\cdots+n_rx_r\ (n_i\in\mathbb{Z},\ 1\leq i\leq r)$$
form a subgroup of $G$, which we denote $\langle x_1,\cdots,x_r\rangle$. $\langle x_1,\cdots,x_r\rangle$ is called a subgroup of $G$ generated by the generators $x_1,\cdots,x_r$. If $G$ itself is generated by finite elements, $G$ is said to be finitely generated. If $n_1x_1+\cdots+n_rx_r=0$ is satisfied only when $n_1=\cdots=n_r=0$, $x_1,\cdots,x_r$ are said to be linearly independent.

Definition. If $G$ is finitely generated by $r$ linearly independent elements, $G$ is called a free abelian group of rank $r$.

Example. $\mathbb{Z}$ is a free abelian group of rank 1 generated by 1 (or $-1$).

Example. Let $\mathbb{Z}\oplus\mathbb{Z}=\{(m,n):m,n\in\mathbb{Z}\}$. The $\mathbb{Z}\oplus\mathbb{Z}$ is a free abelian group of rank 2 generated by $(1,0)$ and $(0,1)$. More generally,
$$\stackrel{r\ \mbox{copies}}{\overbrace{\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}$$
is a free abelian group of rank $r$.

Example. $\mathbb{Z}_2=\{0,1\}$ is finitely generated by 1 but is not free. $1+1=0$ so 1 is not linearly independent.

If $G=\langle x\rangle=\{0,\pm x,\pm 2x,\cdots\}$, $G$ is called a cyclic group. If $nx\ne 0$ $\forall n\in\mathbb{Z}\setminus\{0\}$, it is an infinite cyclic group. If $nx=0$ for some $n\in\mathbb{Z}\setminus\{0\}$, it is a finite cyclic group. Let $G=\langle x\rangle$ and let $f:\mathbb{Z}\longrightarrow G$ be a homomorphism defined by $f(k)=kx$, $k\in\mathbb{Z}$. $f$ is an epimorphism (i.e. onto homomorphism), so by Fundamental Theorem of Homomorphism,
$$G\cong\mathbb{Z}/\ker f.$$
If $G$ is a finite group, then there exists the smallest positive integer $N$ such that $Nx=0$. Thus
$$\ker f=\{0,\pm N,\pm 2N,\cdots\}=N\mathbb{Z}.$$
Hence
$$G\cong\mathbb{Z}/N\mathbb{Z}\cong\mathbb{Z}_N.$$
If $G$ is an infinite cyclic group, $\ker f=\{0\}$. Hence,
$$G\cong\mathbb{Z}/\{0\}\cong\mathbb{Z}.$$

Lemma. Let $G$ be a free abelian group of rank $r$, and let $H\leq G$. Then one may always choose $p$ generators $x_1,\cdots,x_p$ out of $r$ generators of $G$ so that $k_1x_1,\cdots,k_px_p$ generate $H$. Hence,
$$H\cong k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}$$
and $H$ is of rank $p$.

Theorem. [Fundamental Theorem of Finitely Generated Abelian Groups]
Let $G$ be a finitely generated abelian group with $m$ generators. Then
$$G\cong\stackrel{r}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus \mathbb{Z}_{k_1}\oplus\cdots\oplus\mathbb{Z}_{k_p}$$
where $m=r+p$. The number $r$ is called the \emph{rank} of $G$.

Proof. Let $G=\langle x_1, \cdots,x_m\rangle$ and let $f: \mathbb{Z}\oplus\cdots\oplus\mathbb{Z}\longrightarrow G$ be the onto homomorphism
$$f(n_1,\cdots,n_m)=n_1x_1+\cdots +n_mx_m.$$
Then by Fundamental Theorem of Homomorphism
$$\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/\ker f\cong G.$$
$\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}$ is a free abelian group of rank $m$ and $\ker f$ is a subgroup of $\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$, so by the preceding lemma
$$\ker f\cong k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}.$$
Define $\varphi:\stackrel{p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus \cdots\oplus k_p\mathbb{Z}\longrightarrow\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/k_p\mathbb{Z}$ by
$$\varphi((n_1,\cdots,n_p)+k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z})=(n_1+k_1\mathbb{Z},\cdots,n_p+k_p\mathbb{Z}).$$
Then
$$\stackrel{p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}\stackrel{\varphi}{\cong}\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/k_p\mathbb{Z}.$$
Hence,
\begin{align*}
G&\cong\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/\ker f\\
&\cong\stackrel{m}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}/k_1\mathbb{Z}\oplus\cdots\oplus k_p\mathbb{Z}\\
&\cong\stackrel{m-p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus\mathbb{Z}/k_1\mathbb{Z}\oplus\cdots\oplus Z/k_p\mathbb{Z}\\
&\cong\stackrel{m-p}{\overbrace{\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}}}\oplus\mathbb{Z}_{k_1}\oplus\cdots\oplus\mathbb{Z}_{k_p}.
\end{align*}