Before we discuss homology groups, we review some basics of abelian group theory.
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.
Example. For any k\in\mathbb N, k\mathbb Z=\{kn: n\in\mathbb Z\} is a subgroup of \mathbb Z.
Example. \mathbb Z_2=\{0,1\} is not a subgroup of \mathbb Z.
Let H be a subgroup of 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{eqnarray*}[x]&=&\{y\in G: y\sim x\}\\&=&\{y\in G: y-x\in H\}.\end{eqnarray*} 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 1. Let f: G_1\longrightarrow G_2 be a homomorphism. Then
(a) \ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0) is a subgroup of G_1.
(b) {\mathrm im}f=\{f(x): x\in G_1\} is a subgroup of G_2.
Theorem 2 [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 lelements, 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 fintely 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 fintely 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 3. Let G be a free abelian group of rank r, and let H be a subgroup of 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 4 [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 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 surjective 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 Lemma 3 \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{eqnarray*}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{eqnarray*}