To go further with homology, we will need some additional theory of abelian groups.
We will almost always use additive notation for abelian groups, so the group operation will be denoted by $a+b$, the identity element by $0$, and the inverse of $a$ by $-a$. The product of groups $A$ and $B$ will usually be written $A\oplus B$ rather than $A\times B$.
The basic examples of finitely generated abelian groups are $\mathbb{Z}$ and $\mathbb{Z}/n$. Recall that the elements of $\mathbb{Z}/n$ are the cosets $\overline{i}=i+n\mathbb{Z}$. These can be defined for all $i\in \mathbb{Z}$, but they repeat with period $n$, so $\{\overline{0},\mathrm{\dots},\overline{n-1}\}$ is a complete list of elements.
Let $n$, $m$ and $k$ be integers such that $n\mathrm{,}m\mathrm{>}\mathrm{0}$ and $k\mathit{}n$ is divisible by $m$. Then there is a well-defined homomorphism $\varphi \mathrm{:}\mathbb{Z}\mathrm{/}n\mathrm{\to}\mathbb{Z}\mathrm{/}m$ given by $\varphi \mathit{}\mathrm{(}i\mathrm{+}n\mathit{}\mathbb{Z}\mathrm{)}\mathrm{=}i\mathit{}k\mathrm{+}m\mathit{}\mathbb{Z}$.
We can certainly define a homomorphism ${\varphi}_{0}:\mathbb{Z}\to \mathbb{Z}/m$ by ${\varphi}_{0}(i)=ik+m\mathbb{Z}$. By assumption we have $kn=qm$ for some $q$, so ${\varphi}_{0}(in)=ink+m\mathbb{Z}=iqm+m\mathbb{Z}$, and this is the same as $0+m\mathbb{Z}$ because $iqm\in m\mathbb{Z}$. This shows that $n\mathbb{Z}\le \mathrm{ker}({\varphi}_{0})$, so we have a well-defined homomorphism $\varphi $ as described. ∎
Suppose that $n$ and $m$ are positive integers that are coprime. Then there is an isomorphism $\varphi \mathrm{:}\mathbb{Z}\mathrm{/}n\mathit{}m\mathrm{\to}\mathbb{Z}\mathrm{/}n\mathrm{\oplus}\mathbb{Z}\mathrm{/}m$ given by
$$\varphi (i+nm\mathbb{Z})=(i+n\mathbb{Z},i+m\mathbb{Z}).$$ |
It is clear that the above formula gives a well-defined homomorphism. Next, as $n$ and $m$ are coprime, we can find integers $a,b$ such that $an+bm=1$. By the lemma, there are well-defined homomorphisms $\alpha :\mathbb{Z}/n\to \mathbb{Z}/nm$ and $\beta :\mathbb{Z}/m\to \mathbb{Z}/nm$ given by $\alpha (j+n\mathbb{Z})=bmj+nm\mathbb{Z}$ and $\beta (k+m\mathbb{Z})=ank+nm\mathbb{Z}$. We can combine these to define $\psi :\mathbb{Z}/n\oplus \mathbb{Z}/m\to \mathbb{Z}/nm$ by $\psi (u,v)=\alpha (u)+\beta (v)$. Taking account of the identity $an+bm=1$, this can be written more explicitly as
$$\psi (j+n\mathbb{Z},k+m\mathbb{Z})=bmj+ank+nm\mathbb{Z}=j+an(k-j)+nm\mathbb{Z}=k+bm(j-k)+nm\mathbb{Z}.$$ |
From the third and fourth expressions we see that this is equal to $j$ mod $n$ and equal to $k$ mod $m$, so $\varphi \circ \psi =\mathrm{id}$. On the other hand, we have
$$\psi (\varphi (i+nm\mathbb{Z}))=\psi (i+n\mathbb{Z},i+m\mathbb{Z})=bmi+ani+nm\mathbb{Z}=(an+bm)i+nm\mathbb{Z}=i+nm\mathbb{Z},$$ |
so $\psi \circ \varphi $ is also the identity. ∎
Suppose that $n\mathrm{=}{p}_{\mathrm{1}}^{{v}_{\mathrm{1}}}\mathit{}\mathrm{\cdots}\mathit{}{p}_{r}^{{v}_{r}}$, where ${p}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{p}_{r}$ are distinct primes and ${v}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{v}_{r}\mathrm{\ge}\mathrm{0}$. Then there is an isomorphism
$$\varphi :\mathbb{Z}/n\to \mathbb{Z}/{p}_{1}^{{v}_{1}}\oplus \mathrm{\cdots}\oplus \mathbb{Z}/{p}_{r}^{{v}_{r}}$$ |
given by
$$\varphi (i+n\mathbb{Z})=(i+{p}_{1}^{{v}_{1}}\mathbb{Z},\mathrm{\dots},i+{p}_{r}^{{v}_{r}}\mathbb{Z}).$$ |
This follows from Proposition 12.2 by induction on $r$. ∎
Let $A$ be an abelian group, and let $L=({a}_{1},\mathrm{\dots},{a}_{r})$ be a finite list of elements of $A$. Put
$$B=\{{n}_{1}{a}_{1}+\mathrm{\cdots}+{n}_{r}{a}_{r}|{n}_{1},\mathrm{\dots},{n}_{r}\in \mathbb{Z}\}.$$ |
This is easily seen to be a subgroup of $A$. We say that $L$ generates $A$ if $B=A$. We say that $A$ is finitely generated if there is a finite list that generates it.
The list $(1)$ generates $\mathbb{Z}$.
For $r\ge 0$, the standard basis vectors $({e}_{1},\mathrm{\dots},{e}_{r})$ generate ${\mathbb{Z}}^{r}$.
The list $(1+n\mathbb{Z})$ generates $\mathbb{Z}/n$.
If $({a}_{1},\mathrm{\dots},{a}_{r})$ generates $A$ and $({b}_{1},\mathrm{\dots},{b}_{s})$ generates $B$ then the list
$$(({a}_{1},0),\mathrm{\dots},({a}_{r},0),(0,{b}_{1}),\mathrm{\dots},(0,{b}_{s}))$$ |
generates $A\oplus B$. Thus, if $A$ and $B$ are finitely generated, then so is $A\oplus B$.
We can now see by induction that any group of the form
$$A={\mathbb{Z}}^{r}\oplus \mathbb{Z}/{n}_{1}\oplus \mathrm{\cdots}\oplus \mathbb{Z}/{n}_{s}$$ |
is finitely generated.
Suppose again that $({a}_{1},\mathrm{\dots},{a}_{r})$ generates $A$. Let $B$ be a subgroup of $A$, and let $\pi :A\to A/B$ be the usual projection homomorphism, given by $\pi (x)=x+B$. It is then easy to see that the list $(\pi ({a}_{1}),\mathrm{\dots},\pi ({a}_{r}))$ generates $A/B$, so $A/B$ is again finitely generated.
If $A$ is a finite group then it is certainly finitely generated, because we can just take the full list of elements as generators.
Consider the polynomial ring $\mathbb{Z}[x]$ as an abelian group under addition; we will show that this is not finitely generated. Let $L=({f}_{1},\mathrm{\dots},{f}_{r})$ be a finite list of elements of $\mathbb{Z}[x]$, and let $B$ be the subgroup of $\mathbb{Z}$-linear combinations of this list. Let $d$ be the maximum of the degrees of all the polynomials ${f}_{i}$. Any element of $B$ has the form ${\sum}_{i}{n}_{i}{f}_{i}$ for some integers ${n}_{i}$, and so has degree at most $d$. It follows that ${x}^{d+1}\notin B$, so $B$ is not all of $\mathbb{Z}[x]$, so $L$ does not generate $\mathbb{Z}[x]$.
Consider the field $\mathbb{Q}$ as an abelian group under addition; we will show that this is not finitely generated. Let $L=({q}_{1},\mathrm{\dots},{q}_{r})$ be a finite list of elements of $\mathbb{Q}$, and let $B$ be the subgroup of $\mathbb{Z}$-linear combinations of this list. We can write ${q}_{i}$ as ${a}_{i}/{b}_{i}$ for some ${a}_{i},{b}_{i}\in \mathbb{Z}$ with ${b}_{i}>0$. Put $b={b}_{1}{b}_{2}\mathrm{\cdots}{b}_{r}$, so $b{q}_{i}\in \mathbb{Z}$ for all $i$. It follows easily that $bx\in \mathbb{Z}$ for all $x\in B$, so $1/(2b)\notin B$, so $B$ is not all of $\mathbb{Q}$, so $L$ does not generate $\mathbb{Q}$.
We now quote two theorems whose proofs can be found in almost any textbook on abstract algebra.
Let $A$ be a finitely generated abelian group, and let $B$ be a subgroup of $A$. Then $B$ is also finitely generated. ∎
Let $A$ be a finitely generated abelian group. Then $A$ can be expressed (up to isomorphism) as the direct sum of a finite list of summands of the form $\mathbb{Z}$ or $\mathbb{Z}\mathrm{/}{p}^{v}$ (with $p$ prime and $v\mathrm{>}\mathrm{0}$). Moreover, the list of summands is unique up to order. ∎
Put $A={\mathbb{Z}}^{3}$ and $$ and $C=A/B$. We claim that $C\simeq {\mathbb{Z}}^{2}\oplus \mathbb{Z}/2$. Indeed, we can define maps
$${\mathbb{Z}}^{2}\oplus \mathbb{Z}/2\stackrel{\mathit{\varphi}}{\to}C\stackrel{\mathit{\psi}}{\to}{\mathbb{Z}}^{2}\oplus \mathbb{Z}/2$$ |
by
$\varphi (i,j,k+2\mathbb{Z})$ | $=(i+k,j+k,k)+C$ | ||
$\psi ((p,q,r)+C)$ | $=(p-r,q-r,r+2\mathbb{Z})$ |
It is an exercise to check that these are well-defined and inverse to each other.
Consider an abelian group $A$ with $|A|=72={2}^{3}{3}^{2}$. Theorem 12.9 tells us that this can be decomposed as a direct sum of groups of the form $\mathbb{Z}/{p}^{v}$, where $p$ is prime and $v>0$. If $\mathbb{Z}/{p}^{v}$ is a summand, then the order ${p}^{v}$ must divide $|A|$. Thus, the only possible summands are $\mathbb{Z}/2$, $\mathbb{Z}/4$, $\mathbb{Z}/8$, $\mathbb{Z}/3$ and $\mathbb{Z}/9$. The only possibilities for the $2$-power summands are $\mathbb{Z}/8$ or $\mathbb{Z}/2\oplus \mathbb{Z}/4$ or $\mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/2$. The only possibilities for the $3$-power summands are $\mathbb{Z}/9$ or $\mathbb{Z}/3\oplus \mathbb{Z}/3$. Thus, there are six possibilities for $A$:
${A}_{1}$ | $=\mathbb{Z}/8\oplus \mathbb{Z}/9$ | ||
${A}_{2}$ | $=\mathbb{Z}/2\oplus \mathbb{Z}/4\oplus \mathbb{Z}/9$ | ||
${A}_{3}$ | $=\mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/9$ | ||
${A}_{4}$ | $=\mathbb{Z}/8\oplus \mathbb{Z}/3\oplus \mathbb{Z}/3$ | ||
${A}_{5}$ | $=\mathbb{Z}/2\oplus \mathbb{Z}/4\oplus \mathbb{Z}/3\oplus \mathbb{Z}/3$ | ||
${A}_{6}$ | $=\mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/3\oplus \mathbb{Z}/3.$ |
One might think that there were additional possibilities like $\mathbb{Z}/36\oplus \mathbb{Z}/2$, but the Chinese Remainder Theorem gives $\mathbb{Z}/36\simeq \mathbb{Z}/4\oplus \mathbb{Z}/9$, so $\mathbb{Z}/36\oplus \mathbb{Z}/2\simeq {A}_{2}$. Similarly, we have $\mathbb{Z}/72\simeq {A}_{1}$.
The following definition will turn out to be very important.
Consider a pair of homomorphisms $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C$ of abelian groups. Recall that we have subgroups
$\mathrm{img}(\alpha )$ | $=\{\alpha (a)|a\in A\}=\{b\in B|b=\alpha (a)\text{for some}a\in A\}\le B$ | ||
$\mathrm{ker}(\beta )$ | $=\{b\in B|\beta (b)=0\}\le B.$ |
We say that the pair is exact if $\mathrm{img}(\alpha )=\mathrm{ker}(\beta )$. We also say that the pair is short exact if it is exact, and $\alpha $ is injective, and $\beta $ is surjective.
It is standard that $\alpha $ is injective iff $\mathrm{ker}(\alpha )=0$ and $\beta $ is surjective iff $\mathrm{img}(\beta )=C$. Thus, the sequence is short exact iff we have $\mathrm{ker}(\alpha )=0$ and $\mathrm{img}(\alpha )=\mathrm{ker}(\beta )$ and $\mathrm{img}(\beta )=C$.
For any abelian groups $A$ and $B$ we have a short exact sequence $A\stackrel{\mathit{j}}{\to}A\oplus B\stackrel{\mathit{q}}{\to}B$ given by $j(a)=(a,0)$ and $q(a,b)=b$.
For any $n>0$ we have a short exact sequence $\mathbb{Z}\stackrel{\mathit{\mu}}{\to}\mathbb{Z}\stackrel{\mathit{\pi}}{\to}\mathbb{Z}/n$ given by $\mu (x)=nx$ and $\pi (y)=y+n\mathbb{Z}$.
For any $n,m>0$ we have a short exact sequence $\mathbb{Z}/n\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/nm\stackrel{\mathit{\beta}}{\to}\mathbb{Z}/m$ given by $\alpha (i+n\mathbb{Z})=mi+nm\mathbb{Z}$ and $\beta (j+nm\mathbb{Z})=j+m\mathbb{Z}$.
If we define $\alpha :{\mathbb{Z}}^{2}\to {\mathbb{Z}}^{2}$ by $\alpha (x,y)=(y,0)$, then the sequence ${\mathbb{Z}}^{2}\stackrel{\mathit{\alpha}}{\to}{\mathbb{Z}}^{2}\stackrel{\mathit{\alpha}}{\to}{\mathbb{Z}}^{2}$ is exact but not short exact.
A sequence $0\stackrel{}{\to}B\stackrel{\mathit{\beta}}{\to}C$ is exact iff $\beta $ is injective.
A sequence $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{}{\to}0$ is exact iff $\alpha $ is surjective.
If $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C$ is exact, then $\alpha =0$ iff $\beta $ is injective.
If $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C$ is exact, then $\beta =0$ iff $\alpha $ is surjective.
A sequence $0\stackrel{}{\to}A\stackrel{}{\to}0$ is exact iff $A=0$.
Consider a sequence $0\stackrel{}{\to}B\stackrel{}{\to}C$. The image of the first map can only be zero, so the sequence is exact iff $\mathrm{ker}(\beta )=0$, which means that $\beta $ is injective.
Consider a sequence $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{}{\to}0$. The kernel of the second map is all of $B$, so the sequence is exact iff $\mathrm{img}(\alpha )=B$, which means that $\alpha $ is surjective.
Consider an exact sequence $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C$. It is clear that $\alpha $ is the zero homomorphism iff $\mathrm{img}(\alpha )$ is the zero subgroup of $B$, and $\beta $ is injective iff $\mathrm{ker}(\beta )=0$. As $\mathrm{img}(\alpha )$ and $\mathrm{ker}(\beta )$ are the same, we see that $\alpha =0$ iff $\beta $ is injective.
Similarly, it is clear that $\alpha $ is surjective iff $\mathrm{img}(\alpha )=B$, and $\beta =0$ iff $\mathrm{ker}(\beta )=B$. As $\mathrm{img}(\alpha )$ and $\mathrm{ker}(\beta )$ are the same, we see that $\alpha $ is surjective iff $\beta =0$.
Consider a sequence $0\stackrel{}{\to}A\stackrel{}{\to}0$. The image of the first map is $0$, and the kernel of the second map is $A$. The sequence is exact iff these two subgroups are the same, iff $A=0$.
∎
Consider a longer sequence
$$\mathrm{\cdots}{A}_{i-2}\stackrel{{\alpha}_{i-2}}{\to}{A}_{i-1}\stackrel{{\alpha}_{i-1}}{\to}{A}_{i}\stackrel{{\alpha}_{i}}{\to}{A}_{i+1}\stackrel{{\alpha}_{i+1}}{\to}{A}_{i+2}\mathrm{\cdots}$$ |
We say that this is exact at ${A}_{i}$ if $\mathrm{img}({\alpha}_{i-1})=\mathrm{ker}({\alpha}_{i})$, so the subsequence
$${A}_{i-1}\stackrel{{\alpha}_{i-1}}{\to}{A}_{i}\stackrel{{\alpha}_{i}}{\to}{A}_{i+1}$$ |
is exact. We say that the whole sequence is exact if it is exact at ${A}_{i}$ for all $i$.
Consider a sequence
$$0\stackrel{}{\to}A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C\stackrel{}{\to}0.$$ |
This is exact at $A$ iff $\alpha $ is injective, and exact at $C$ iff $\beta $ is surjective. Thus, the whole sequence is exact iff $A\stackrel{\mathit{\alpha}}{\to}B\stackrel{\mathit{\beta}}{\to}C$ is short exact.
Consider the group $\mathbb{Z}/4=\{0,1,2,3\}$ and the homomorphism $\alpha :\mathbb{Z}/4\to \mathbb{Z}/4$ given by $\alpha (a)=2a$. We then find that $\mathrm{img}(\alpha )=\mathrm{ker}(\alpha )=\{0,2\}$, so the sequence
$$\mathrm{\cdots}\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathbb{Z}/4\stackrel{\mathit{\alpha}}{\to}\mathrm{\cdots}$$ |
is exact.
A sequence $\mathrm{0}\stackrel{}{\mathrm{\to}}A\stackrel{\mathit{\alpha}}{\mathrm{\to}}B\stackrel{}{\mathrm{\to}}\mathrm{0}$ is exact iff $\alpha $ is an isomorphism.
The sequence is exact iff $\mathrm{ker}(\alpha )$ is the image of the map $0\to A$ (so $\mathrm{ker}(\alpha )=0$) and $\mathrm{img}(\alpha )$ is the kernel of the map $B\to 0$ (which is all of $B$). The first condition means that $\alpha $ is injective, and the second means that $\alpha $ is surjective, so the two conditions together mean that $\alpha $ is an isomorphism. ∎
Let $A\stackrel{\mathit{\alpha}}{\mathrm{\to}}B\stackrel{\mathit{\beta}}{\mathrm{\to}}C$ be a short exact sequence. Then $A$ is isomorphic to the subgroup $\mathrm{img}\mathit{}\mathrm{(}\alpha \mathrm{)}\mathrm{=}\alpha \mathit{}\mathrm{(}A\mathrm{)}\mathrm{\le}B$, and the quotient group $B\mathrm{/}\alpha \mathit{}\mathrm{(}A\mathrm{)}$ is isomorphic to $C$. Moreover, the group $B$ is finite iff $A$ and $C$ are both finite, and if so, we have $\mathrm{|}B\mathrm{|}\mathrm{=}\mathrm{|}A\mathrm{|}\mathit{}\mathrm{|}C\mathrm{|}$.
As the sequence is assumed to be short exact, we know that $\alpha $ is injective and $\beta $ is surjective and $\mathrm{img}(\alpha )=\mathrm{ker}(\beta )$.
We can considar $\alpha $ as a homomorphism from $A$ to $\alpha (A)$. In this context it is surjective (by definition of $\alpha (A)$) and injective (by assumption) so it is an isomorphism, as required.
Next, as $\beta :B\to C$ is a surjective homomorphism, the First Isomorphism Theorem tells us that it induces an isomorphism $B/\mathrm{ker}(\beta )\to C$. By the exactness assumption we have $\mathrm{ker}(\beta )=\alpha (A)$, so we have $B/\alpha (A)\simeq C$ as claimed.
Suppose that $A$ and $C$ are finite. We can divide $B$ into cosets for the subgroup $\alpha (A)$. As $\alpha :A\to \alpha (A)$ is an isomorphism we have $|\alpha (A)|=|A|$, so each coset has size $|A|$. There is one coset for each element of the group $B/\alpha (A)\simeq C$, so the number of cosets is $|C|$. Thus, the total number of elements is $$.
Conversely, suppose that $B$ is finite. As $A$ is isomorphic to a subgroup of $B$, it is also finite. As $C$ is isomorphic to a quotient of $B$, it is also finite. We can then go back to the last paragraph to see that $|B|=|A||C|$ again. ∎