# 12. Abelian groups

To go further with homology, we will need some additional theory of abelian groups.

Video (Lemma 12.1 to Corollary 12.3)

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 $\{\overline{0},\dotsc,\overline{n-1}\}$ is a complete list of elements.

###### Lemma 12.1.

Let $n$, $m$ and $k$ be integers such that $n,m>0$ and $kn$ is divisible by $m$. Then there is a well-defined homomorphism $\phi\colon{\mathbb{Z}}/n\to{\mathbb{Z}}/m$ given by $\phi(i+n{\mathbb{Z}})=ik+m{\mathbb{Z}}$.

###### Proof.

We can certainly define a homomorphism $\phi_{0}\colon{\mathbb{Z}}\to{\mathbb{Z}}/m$ by $\phi_{0}(i)=ik+m{\mathbb{Z}}$. By assumption we have $kn=qm$ for some $q$, so $\phi_{0}(in)=ink+m{\mathbb{Z}}=iqm+m{\mathbb{Z}}$, and this is the same as $0+m{\mathbb{Z}}$ because $iqm\in m{\mathbb{Z}}$. This shows that $n{\mathbb{Z}}\leq\ker(\phi_{0})$, so we have a well-defined homomorphism $\phi$ as described. ∎

###### Proposition 12.2(The Chinese Remainder Theorem).

Suppose that $n$ and $m$ are positive integers that are coprime. Then there is an isomorphism $\phi\colon{\mathbb{Z}}/nm\to{\mathbb{Z}}/n\oplus{\mathbb{Z}}/m$ given by

 $\phi(i+nm{\mathbb{Z}})=(i+n{\mathbb{Z}},\;i+m{\mathbb{Z}}).$
###### Proof.

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 $\alpha\colon{\mathbb{Z}}/n\to{\mathbb{Z}}/nm$ and $\beta\colon{\mathbb{Z}}/m\to{\mathbb{Z}}/nm$ 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 $\psi\colon{\mathbb{Z}}/n\oplus{\mathbb{Z}}/m\to{\mathbb{Z}}/nm$ 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 $\phi\circ\psi=\operatorname{id}$. On the other hand, we have

 $\psi(\phi(i+nm{\mathbb{Z}}))=\psi(i+n{\mathbb{Z}},i+m{\mathbb{Z}})=bmi+ani+nm{% \mathbb{Z}}=(an+bm)i+nm{\mathbb{Z}}=i+nm{\mathbb{Z}},$

so $\psi\circ\phi$ is also the identity. ∎

###### Corollary 12.3.

Suppose that $n=p_{1}^{v_{1}}\dotsb p_{r}^{v_{r}}$, where $p_{1},\dotsc,p_{r}$ are distinct primes and $v_{1},\dotsc,v_{r}\geq 0$. Then there is an isomorphism

 $\phi\colon{\mathbb{Z}}/n\to{\mathbb{Z}}/p_{1}^{v_{1}}\oplus\dotsb\oplus{% \mathbb{Z}}/p_{r}^{v_{r}}$

given by

 $\phi(i+n{\mathbb{Z}})=(i+p_{1}^{v_{1}}{\mathbb{Z}},\dotsc,i+p_{r}^{v_{r}}{% \mathbb{Z}}).$
###### Proof.

This follows from Proposition 12.2 by induction on $r$. ∎

Video (Definition 12.4 to Example 12.11)

###### Definition 12.4.

Let $A$ be an abelian group, and let $L=(a_{1},\dotsc,a_{r})$ be a finite list of elements of $A$. Put

 $B=\{n_{1}a_{1}+\dotsb+n_{r}a_{r}\;|\;n_{1},\dotsc,n_{r}\in{\mathbb{Z}}\}.$

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.

###### Example 12.5.
• (a)

The list $(1)$ generates ${\mathbb{Z}}$.

• (b)

For $r\geq 0$, the standard basis vectors $(e_{1},\dotsc,e_{r})$ generate ${\mathbb{Z}}^{r}$.

• (c)

The list $(1+n{\mathbb{Z}})$ generates ${\mathbb{Z}}/n$.

• (d)

If $(a_{1},\dotsc,a_{r})$ generates $A$ and $(b_{1},\dotsc,b_{s})$ generates $B$ then the list

 $((a_{1},0),\dotsc,(a_{r},0),\;(0,b_{1}),\dotsc,(0,b_{s}))$

generates $A\oplus B$. Thus, if $A$ and $B$ are finitely generated, then so is $A\oplus B$.

• (e)

We can now see by induction that any group of the form

 $A={\mathbb{Z}}^{r}\oplus{\mathbb{Z}}/n_{1}\oplus\dotsb\oplus{\mathbb{Z}}/n_{s}$

is finitely generated.

• (f)

Suppose again that $(a_{1},\dotsc,a_{r})$ generates $A$. Let $B$ be a subgroup of $A$, and let $\pi\colon A\to A/B$ be the usual projection homomorphism, given by $\pi(x)=x+B$. It is then easy to see that the list $(\pi(a_{1}),\dotsc,\pi(a_{r}))$ generates $A/B$, so $A/B$ is again finitely generated.

• (g)

If $A$ is a finite group then it is certainly finitely generated, because we can just take the full list of elements as generators.

###### Example 12.6.

Consider the polynomial ring ${\mathbb{Z}}[x]$ as an abelian group under addition; we will show that this is not finitely generated. Let $L=(f_{1},\dotsc,f_{r})$ 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 $x^{d+1}\not\in B$, so $B$ is not all of ${\mathbb{Z}}[x]$, so $L$ does not generate ${\mathbb{Z}}[x]$.

###### Example 12.7.

Consider the field ${\mathbb{Q}}$ as an abelian group under addition; we will show that this is not finitely generated. Let $L=(q_{1},\dotsc,q_{r})$ 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 $b=b_{1}b_{2}\dotsb b_{r}$, so $bq_{i}\in{\mathbb{Z}}$ for all $i$. It follows easily that $bx\in{\mathbb{Z}}$ for all $x\in B$, so $1/(2b)\not\in B$, 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.

###### Proposition 12.8.

Let $A$ be a finitely generated abelian group, and let $B$ be a subgroup of $A$. Then $B$ is also finitely generated. ∎

###### Theorem 12.9.

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 ${\mathbb{Z}}/p^{v}$ (with $p$ prime and $v>0$). Moreover, the list of summands is unique up to order. ∎

###### Example 12.10.

Put $A={\mathbb{Z}}^{3}$ and $B=\{(2n,2n,2n)\;|\;n\in{\mathbb{Z}}\} and $C=A/B$. We claim that $C\simeq{\mathbb{Z}}^{2}\oplus{\mathbb{Z}}/2$. Indeed, we can define maps

 ${\mathbb{Z}}^{2}\oplus{\mathbb{Z}}/2\xrightarrow{\phi}C\xrightarrow{\psi}{% \mathbb{Z}}^{2}\oplus{\mathbb{Z}}/2$

by

 $\displaystyle\phi(i,j,k+2{\mathbb{Z}})$ $\displaystyle=(i+k,j+k,k)+C$ $\displaystyle\psi((p,q,r)+C)$ $\displaystyle=(p-r,q-r,r+2{\mathbb{Z}})$

It is an exercise to check that these are well-defined and inverse to each other.

###### Example 12.11.

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 $A$:

 $\displaystyle A_{1}$ $\displaystyle={\mathbb{Z}}/8\oplus{\mathbb{Z}}/9$ $\displaystyle A_{2}$ $\displaystyle={\mathbb{Z}}/2\oplus{\mathbb{Z}}/4\oplus{\mathbb{Z}}/9$ $\displaystyle A_{3}$ $\displaystyle={\mathbb{Z}}/2\oplus{\mathbb{Z}}/2\oplus{\mathbb{Z}}/2\oplus{% \mathbb{Z}}/9$ $\displaystyle A_{4}$ $\displaystyle={\mathbb{Z}}/8\oplus{\mathbb{Z}}/3\oplus{\mathbb{Z}}/3$ $\displaystyle A_{5}$ $\displaystyle={\mathbb{Z}}/2\oplus{\mathbb{Z}}/4\oplus{\mathbb{Z}}/3\oplus{% \mathbb{Z}}/3$ $\displaystyle A_{6}$ $\displaystyle={\mathbb{Z}}/2\oplus{\mathbb{Z}}/2\oplus{\mathbb{Z}}/2\oplus{% \mathbb{Z}}/3\oplus{\mathbb{Z}}/3.$

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 ${\mathbb{Z}}/72\simeq A_{1}$.

Video (Definition 12.12 to Lemma 12.20)

The following definition will turn out to be very important.

###### Definition 12.12.

Consider a pair of homomorphisms $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$ of abelian groups. Recall that we have subgroups

 $\displaystyle\operatorname{img}(\alpha)$ $\displaystyle=\{\alpha(a)\;|\;a\in A\}=\{b\in B\;|\;b=\alpha(a)\text{ for some% }a\in A\}\leq B$ $\displaystyle\ker(\beta)$ $\displaystyle=\{b\in B\;|\;\beta(b)=0\}\leq B.$

We say that the pair is exact if $\operatorname{img}(\alpha)=\ker(\beta)$. We also say that the pair is short exact if it is exact, and $\alpha$ is injective, and $\beta$ is surjective.

###### Remark 12.13.

It is standard that $\alpha$ is injective iff $\ker(\alpha)=0$ and $\beta$ is surjective iff $\operatorname{img}(\beta)=C$. Thus, the sequence is short exact iff we have $\ker(\alpha)=0$ and $\operatorname{img}(\alpha)=\ker(\beta)$ and $\operatorname{img}(\beta)=C$.

###### Example 12.14.
• (a)

For any abelian groups $A$ and $B$ we have a short exact sequence $A\xrightarrow{j}A\oplus B\xrightarrow{q}B$ given by $j(a)=(a,0)$ and $q(a,b)=b$.

• (b)

For any $n>0$ we have a short exact sequence ${\mathbb{Z}}\xrightarrow{\mu}{\mathbb{Z}}\xrightarrow{\pi}{\mathbb{Z}}/n$ given by $\mu(x)=nx$ and $\pi(y)=y+n{\mathbb{Z}}$.

• (c)

For any $n,m>0$ we have a short exact sequence ${\mathbb{Z}}/n\xrightarrow{\alpha}{\mathbb{Z}}/nm\xrightarrow{\beta}{\mathbb{Z% }}/m$ given by $\alpha(i+n{\mathbb{Z}})=mi+nm{\mathbb{Z}}$ and $\beta(j+nm{\mathbb{Z}})=j+m{\mathbb{Z}}$.

• (d)

If we define $\alpha\colon{\mathbb{Z}}^{2}\to{\mathbb{Z}}^{2}$ by $\alpha(x,y)=(y,0)$, then the sequence ${\mathbb{Z}}^{2}\xrightarrow{\alpha}{\mathbb{Z}}^{2}\xrightarrow{\alpha}{% \mathbb{Z}}^{2}$ is exact but not short exact.

###### Proposition 12.15.
• (a)

A sequence $0\xrightarrow{}B\xrightarrow{\beta}C$ is exact iff $\beta$ is injective.

• (b)

A sequence $A\xrightarrow{\alpha}B\xrightarrow{}0$ is exact iff $\alpha$ is surjective.

• (c)

If $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$ is exact, then $\alpha=0$ iff $\beta$ is injective.

• (d)

If $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$ is exact, then $\beta=0$ iff $\alpha$ is surjective.

• (e)

A sequence $0\xrightarrow{}A\xrightarrow{}0$ is exact iff $A=0$.

###### Proof.
• (a)

Consider a sequence $0\xrightarrow{}B\xrightarrow{}C$. The image of the first map can only be zero, so the sequence is exact iff $\ker(\beta)=0$, which means that $\beta$ is injective.

• (b)

Consider a sequence $A\xrightarrow{\alpha}B\xrightarrow{}0$. The kernel of the second map is all of $B$, so the sequence is exact iff $\operatorname{img}(\alpha)=B$, which means that $\alpha$ is surjective.

• (c)

Consider an exact sequence $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$. It is clear that $\alpha$ is the zero homomorphism iff $\operatorname{img}(\alpha)$ is the zero subgroup of $B$, and $\beta$ is injective iff $\ker(\beta)=0$. As $\operatorname{img}(\alpha)$ and $\ker(\beta)$ are the same, we see that $\alpha=0$ iff $\beta$ is injective.

• (d)

Similarly, it is clear that $\alpha$ is surjective iff $\operatorname{img}(\alpha)=B$, and $\beta=0$ iff $\ker(\beta)=B$. As $\operatorname{img}(\alpha)$ and $\ker(\beta)$ are the same, we see that $\alpha$ is surjective iff $\beta=0$.

• (e)

Consider a sequence $0\xrightarrow{}A\xrightarrow{}0$. 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 $A=0$.

###### Definition 12.16.

Consider a longer sequence

 $\dotsb A_{i-2}\xrightarrow{\alpha_{i-2}}A_{i-1}\xrightarrow{\alpha_{i-1}}A_{i}% \xrightarrow{\alpha_{i}}A_{i+1}\xrightarrow{\alpha_{i+1}}A_{i+2}\dotsb$

We say that this is exact at $A_{i}$ if $\operatorname{img}(\alpha_{i-1})=\ker(\alpha_{i})$, so the subsequence

 $A_{i-1}\xrightarrow{\alpha_{i-1}}A_{i}\xrightarrow{\alpha_{i}}A_{i+1}$

is exact. We say that the whole sequence is exact if it is exact at $A_{i}$ for all $i$.

###### Example 12.17.

Consider a sequence

 $0\xrightarrow{}A\xrightarrow{\alpha}B\xrightarrow{\beta}C\xrightarrow{}0.$

This is exact at $A$ iff $\alpha$ is injective, and exact at $C$ iff $\beta$ is surjective. Thus, the whole sequence is exact iff $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$ is short exact.

###### Example 12.18.

Consider the group ${\mathbb{Z}}/4=\{0,1,2,3\}$ and the homomorphism $\alpha\colon{\mathbb{Z}}/4\to{\mathbb{Z}}/4$ given by $\alpha(a)=2a$. We then find that $\operatorname{img}(\alpha)=\ker(\alpha)=\{0,2\}$, so the sequence

 $\dotsb\xrightarrow{\alpha}{\mathbb{Z}}/4\xrightarrow{\alpha}{\mathbb{Z}}/4% \xrightarrow{\alpha}{\mathbb{Z}}/4\xrightarrow{\alpha}{\mathbb{Z}}/4% \xrightarrow{\alpha}{\mathbb{Z}}/4\xrightarrow{\alpha}{\mathbb{Z}}/4% \xrightarrow{\alpha}\dotsb$

is exact.

###### Lemma 12.19.

A sequence $0\xrightarrow{}A\xrightarrow{\alpha}B\xrightarrow{}0$ is exact iff $\alpha$ is an isomorphism.

###### Proof.

The sequence is exact iff $\ker(\alpha)$ is the image of the map $0\to A$ (so $\ker(\alpha)=0$) and $\operatorname{img}(\alpha)$ 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. ∎

###### Lemma 12.20.

Let $A\xrightarrow{\alpha}B\xrightarrow{\beta}C$ be a short exact sequence. Then $A$ is isomorphic to the subgroup $\operatorname{img}(\alpha)=\alpha(A)\leq B$, and the quotient group $B/\alpha(A)$ is isomorphic to $C$. Moreover, the group $B$ is finite iff $A$ and $C$ are both finite, and if so, we have $|B|=|A||C|$.

###### Proof.

As the sequence is assumed to be short exact, we know that $\alpha$ is injective and $\beta$ is surjective and $\operatorname{img}(\alpha)=\ker(\beta)$.

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 $\beta\colon B\to C$ is a surjective homomorphism, the First Isomorphism Theorem tells us that it induces an isomorphism $B/\ker(\beta)\to C$. By the exactness assumption we have $\ker(\beta)=\alpha(A)$, 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 $\alpha\colon A\to\alpha(A)$ 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 $|B|=|A||C|<\infty$.

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. ∎