3.1 Abelian groups

The mathematical structures underlying homology groups are finitely generated Abelian groups. Throughout this chapter, the group operation is denoted by $+$ since all the groups considered here are Abelian (commutative). The unit element is denoted by 0.

3.1.1 Elementary group theory

Let $G_1$ and $G_2$ be Abelian groups. A map $f:G_1\to G_2$ is said to be a homomorphism if

$$f(x+y)=f(x)+f(y)\text{\hspace{1em}(3.1)}$$

for any $x,y\in G_1$ . If $f$ is also a bijection, $f$ is called an isomorphism. If there exists an isomorphism $f:G_1\to G_2$ , $G_1$ is said to be isomorphic to $G_2$ , denoted by $G_1\cong G_2$ . For example, a map $f:\mathbb{Z}\to {\mathbb{Z}}_2=\{0,1\}$ defined by

$$f(2n)=0f(2n+1)=1$$

is a homomorphism. Indeed

$$f(2m+2n)=f(2(m+n))=0=0+0=f(2m)+f(2n)$$

$$\begin{array}{rl}f(2m+1+2n+1)& =f(2(m+n+1))=0=1+1\\ & =f(2m+1)+f(2n+1)\end{array}$$

$$\begin{array}{rl}f(2m+1+2n)& =f(2(m+n)+1)=1=1+0\\ & =f(2m+1)+f(2n).\end{array}$$

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$ . For example,

$$k\mathbb{Z}\equiv \{kn|n\in \mathbb{Z}\}k\in \mathbb{N}$$

is a subgroup of $\mathbb{Z}$ , while ${\mathbb{Z}}_2=\{0,1\}$ is not.

Let $H$ be a subgroup of $G$ . We say $x,y\in G$ are equivalent if

$$x-y\in H\text{\hspace{1em}(3.2)}$$

and write $x\sim y$ . Clearly $\sim$ is an equivalence relation. The equivalence class to which $x$ belongs is denoted by $[x]$ . Let $G/H$ be the quotient space. The group operation $+$ in $G$ naturally induces the group operation $+$ in $G/H$ by

$$[x]+[y]=[x+y].\text{\hspace{1em}(3.3)}$$

Note that $+$ on the LHS is an operation in $G/H$ while $+$ on the RHS is that in $G$ . The operation in $G/H$ should be independent of the choice of representatives. In fact, if $[x^{\prime }]=[x]$ , $[y^{\prime }]=[y]$ , then $x-x^{\prime }=h$ , $y-y^{\prime }=g$ for some $h,g\in H$ and we find that

$$x^{\prime }+y^{\prime }=x+y-(h+g)\in [x+y]$$

Furthermore, $G/H$ becomes a group with this operation, since $H$ is always a normal subgroup of $G$ ; see example 2.6. The unit element of $G/H$ is $[0]=[h]$ ,

$h\in H$ . If $H=G$ , $0-x\in G$ for any $x\in G$ and $G/G$ has just one element [0]. If $H=\{0\}$ , $G/H$ is $G$ itself since $x-y=0$ if and only if $x=y$ .

Example 3.1

Let us work out the quotient group $\mathbb{Z}/2\mathbb{Z}$ . For even numbers we have $2n-2m=2(n-m)\in 2\mathbb{Z}$ and $[2m]=[2n]$ . For odd numbers $(2n+1)-(2m+1)=2(n-m)\in 2\mathbb{Z}$ and $[2m+1]=[2n+1]$ . Even numbers and odd numbers never belong to the same equivalence class since $2n-(2m+1)\notin 2\mathbb{Z}$ . Thus, it follows that

$$\mathbb{Z}/2\mathbb{Z}=\{[0],[1]\}.\text{\hspace{1em}(3.4)}$$

If we define an isomorphism $\phi :\mathbb{Z}/2\mathbb{Z}\to {\mathbb{Z}}_2$ by $\phi ([0])=0$ and $\phi ([1])=1$ , we find $\mathbb{Z}/2\mathbb{Z}\cong {\mathbb{Z}}_2$ . For general $k\in \mathbb{N}$ , we have

$$\mathbb{Z}/k\mathbb{Z}\cong {\mathbb{Z}}_k.\text{\hspace{1em}(3.5)}$$

Lemma 3.1

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

(a) $\ker f=\{x|x\in G_1,f(x)=0\}$ is a subgroup of $G_1$
(b) $\mathrm{im}f=\{x|x\in f(G_1)\subset G_2\}$ is a subgroup of $G_2$

Proof

(a) Let $x,y\in \ker f$ . Then $x+y\in \ker f$ since $f(x+y)=f(x)+f(y)=0+0=0$ . Note that $0\in \ker f$ for $f(0)=f(0)+f(0)$ . We also have $-x\in \ker f$ since $f(0)=f(x-x)=f(x)+f(-x)=0$ .

(b) Let $y_1=f(x_1),y_2=f(x_2)\in \mathrm{im}f$ where $x_1,x_2\in G_1$ . Since $f$ is a homomorphism we have $y_1+y_2=f(x_1)+f(x_2)=f(x_1+x_2)\in \mathrm{im}f$ . Clearly $0\in \mathrm{im}f$ since $f(0)=0$ . If $y=f(x),-y\in \mathrm{im}f$ since $0=f(x-x)=f(x)+f(-x)$ implies $f(-x)=-y$ .

Theorem 3.1 · Fundamental theorem of homomorphism

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

$$G_1/\ker f\cong \mathrm{im}f.\text{\hspace{1em}(3.6)}$$

Proof

Both sides are groups according to lemma 3.1. Define a map $\phi :G_1/\ker f\to \mathrm{im}f$ by $\phi ([x])=f(x)$ . This map is well defined since for $x^{\prime }\in [x]$ , there exists $h\in \ker f$ such that $x^{\prime }=x+h$ and $f(x^{\prime })=f(x+h)=f(x)+f(h)=f(x)$ . Now we show that $\phi$ is an isomorphism. First, $\phi$ is a homomorphism,

$$\begin{array}{rl}\phi ([x]+[y])& =\phi ([x+y])=f(x+y)\\ & =f(x)+f(y)=\phi ([x])+\phi ([y]).\end{array}$$

Second, $\phi$ is one to one: if $\phi ([x])=\phi ([y])$ , then $f(x)=f(y)$ or $f(x)-f(y)=f(x-y)=0$ . This shows that $x-y\in \ker f$ and $[x]=[y]$ . Finally, $\phi$ is onto: if $y\in \mathrm{im}f$ , there exists $x\in G_1$ such that $f(x)=y=\phi ([x])$ .

Example 3.2

Let $f:\mathbb{Z}\to {\mathbb{Z}}_2$ be defined by $f(2n)=0$ and $f(2n+1)=1$ . Then $\ker f=2\mathbb{Z}$ and $\mathrm{im}f={\mathbb{Z}}_2$ are groups. Theorem 3.1 states that $\mathbb{Z}/2\mathbb{Z}\cong {\mathbb{Z}}_2$ , in agreement with example 3.1.

3.1.2 Finitely generated Abelian groups and free Abelian groups

Let $x$ be an element of a group $G$ . For $n\in \mathbb{Z},nx$ denotes

$$\underset{n}{\underbrace{x+\cdots +x}}(\text{i f}n>0)$$

and

$$\underset{|n|}{\underbrace{(-x)+\cdots +(-x)}}(\text{i f}n<0).$$

If $n=0$ , we put $0x=0$ . Take $r$ elements $x_1,\dots ,x_r$ of $G$ . The elements of $G$ of the form

$$n_1{x}_1+\cdots +n_r{x}_r(n_i\in \mathbb{Z},1\le i\le r)\text{\hspace{1em}(3.7)}$$

form a subgroup of $G$ , which we denote $H$ . $H$ is called a subgroup of $G$ generated by the generators $x_1,\dots ,x_r$ . If $G$ itself is generated by finite elements $x_1,\dots ,x_r$ , $G$ is said to be finitely generated. If $n_1{x}_1+\cdots +n_r{x}_r=0$ is satisfied only when $n_1=\cdots =n_r=0$ , $x_1,\dots ,x_r$ are said to be linearly independent.

Definition 3.1

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

Example 3.3

$\mathbb{Z}$ is a free Abelian group of rank 1 finitely generated by 1 (or $-1$ ). Let $\mathbb{Z}\oplus \mathbb{Z}$ be the set of pairs $\{(i,j)|i,j\in \mathbb{Z}\}$ . It is a free Abelian group of rank 2 finitely generated by generators $(1,0)$ and $(0,1)$ . More generally

$$\underset{r}{\underbrace{\mathbb{Z}\oplus \mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}$$

is a free Abelian group of rank $r$ . The group ${\mathbb{Z}}_2=\{0,1\}$ is finitely generated by 1 but is not free since 1 is not linearly independent (note $1+1=0$ ).

3.1.3 Cyclic groups

If $G$ is generated by one element $x$ , $G=\{0,\pm x,\pm 2x,\dots \}$ , $G$ is called a cyclic group. If $nx\ne 0$ for any $n\in \mathbb{Z}-\{0\}$ , it is an infinite cyclic group while if $nx=0$ for some $n\in \mathbb{Z}-\{0\}$ , a finite cyclic group. Let $G$ be a cyclic group generated by $x$ and let $f:\mathbb{Z}\to G$ be a homomorphism defined by $f(n)=nx$ . $f$ maps $\mathbb{Z}$ onto $G$ but not necessarily one to one. From theorem 3.1, we have $G=\mathrm{im}f\cong \mathbb{Z}/\ker f$ . Let $N$ be the smallest positive integer such that $Nx=0$ . Clearly

$$\ker f=\{0,\pm N,\pm 2N,\dots \}=N\mathbb{Z}\text{\hspace{1em}(3.8)}$$

and we have

$$G\cong \mathbb{Z}/N\mathbb{Z}\cong {\mathbb{Z}}_N.\text{\hspace{1em}(3.9)}$$

If $G$ is an infinite cyclic group, then $\ker f=\{0\}$ and $G\cong \mathbb{Z}$ . Any infinite cyclic group is isomorphic to $\mathbb{Z}$ while a finite cyclic group is isomorphic to some ${\mathbb{Z}}_N$ .

We will need the following lemma and theorem in due course. We first state the lemma without proof.

Lemma 3.2

Let $G$ be a free Abelian group of rank $r$ and let $H(\ne \varnothing )$ be a subgroup of $G$ . We may always choose $p$ generators $x_1,\dots ,x_p$ , out of $r$ generators of $G$ so that $k_1{x}_1,\dots ,k_p{x}_p$ generate $H$ . Thus, $H\cong k_1\mathbb{Z}\oplus \cdots \oplus k_p\mathbb{Z}$ and $H$ is of rank $p$ .

Theorem 3.2 · Fundamental theorem of finitely generated Abelian groups

Let $G$ be a finitely generated Abelian group (not necessarily free) with $m$ generators. Then $G$ is isomorphic to the direct sum of cyclic groups,

$$G\cong \underset{r}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}\oplus {\mathbb{Z}}_{k_1}\oplus \cdots \oplus {\mathbb{Z}}_{k_p}\text{\hspace{1em}(3.10)}$$

where $m=r+p$ . The number $r$ is called the rank of $G$ .

Proof

Let $G$ be generated by $m$ elements $x_1,\dots ,x_m$ and let

$$f:\underset{m}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}\to G$$

be a surjective homomorphism,

$$f(n_1,\dots ,n_m)=n_1{x}_1+\cdots +n_m{x}_m.$$

Theorem 3.1 states that

$$\underset{m}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}/\ker f\cong G.$$

Since $\ker f$ is a subgroup of

$$\underset{m}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}$$

lemma 3.2 claims that if we choose the generators properly, we have

$$\ker f\cong k_1\mathbb{Z}\oplus \cdots \oplus k_p\mathbb{Z}.$$

We finally obtain

$$\begin{array}{rl}& G\cong \underset{m}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}/\ker f\cong \underset{m}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}/(k_1\mathbb{Z}\oplus \cdots \oplus k_p\mathbb{Z})\\ & \cong \underset{m-p}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}\oplus {\mathbb{Z}}_{k_1}\oplus \cdots \oplus {\mathbb{Z}}_{k_p}.\end{array}$$

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