LEC ALG2 14

Free abelian groups

Free abelian groups are obtained by stating the universal property for free groups in the category $\text{Ab}$.

Definition 103.1(Definition).

Given a non-empty set $S$ and a map $\theta :S\to F^{ab}$ into an abelian group $F^{ab}$, the pair $(F^{ab},\theta )$ is said to be a free abelian group on the set $S$ if, for any function $\phi :S\to G$ to any abelian group $G$, there is a unique homomorphism $\overline{\phi }:F^{ab}\to G$ such that $\phi =\overline{\phi }\circ \theta$. When $\theta$ is an inclusion, we call $\overline{\phi }$ the unique extension of $\phi$ to $F^{ab}$.

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Let’s first tackle the finite case. Denote by ${\mathbb{Z}}^{\oplus n}$ the direct sum

$$\underset{n\text{-times}}{\underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}.$$

Recall that this group is the same as the product ${\mathbb{Z}}^n$ (however, it will be playing the role of a coproduct in what follows).

Proposition 103.2(Proposition).

For $S=\{1,2,\dots ,n\}$, $({\mathbb{Z}}^{\oplus n},\theta )$ is a free abelian group on $S$, where $\theta (k)=(0,\dots ,0,1,0,\dots ,0)$ (the $1$ is in the $k$th coordinate).

Proof.

Let $\phi :S\to G$ be given. For $k\in S$, let ${\phi }_k:\mathbb{Z}\to G$ be defined by $1\mapsto \phi (k)$. Let ${\iota }_i:\mathbb{Z}\to {\mathbb{Z}}^{\oplus n}$ be the coproduct injections. Then, by the universal property of coproducts, there exists a unique map $\Phi :{\mathbb{Z}}^{\oplus n}\to G$ such that $\Phi \circ {\iota }_k={\phi }_k$ for all $1⩽k⩽n$.

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Now, let $S$ be any set. Recall that $H^S={\text{Hom}}_{\text{Set}}(S,H)$ has a natural abelian group structure if $H$ is an abelian group; elements of $H^S$ are arbitrary set-functions $\alpha :S\to H$. Recall that in the general (non-finite) case, the coproduct is the subgroup of the product consisting of tuples with finite support.

Thus, the coproduct ${⨁}_{s\in S}H$, which we will denote by $H^{\oplus S}$, is given by

$$H^{\oplus S}:=\{\alpha :S\to H:\alpha (s)\ne e_H\text{ for only finitely many elements }s\in S\}.$$

For $H=\mathbb{Z}$ the coproduct injections $\theta :S\to {\mathbb{Z}}^{\oplus S}$ are obtained by mapping $s\in S$ to the function ${\theta }_s:S\to \mathbb{Z}$ defined by

$${\theta }_s(x):=\{\begin{array}{ll}1& x=s\\ 0& x\ne s.\end{array}$$

Proposition 103.3(Proposition).

For every set $S$, ${\mathbb{Z}}^{\oplus S}$ is free abelian on $S$.

Proof.

Same as that of Prp 2; just use the universal property of coproducts.□

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Proposition 103.4(Proposition).

1. If $|X_1|=|X_2|$, then $F^{ab}(X_1)\cong F^{ab}(X_2)$.
2. If $F^{ab}(X_1)\cong F^{ab}(X_2)$, then $|X_1|=|X_2|$.

Thus, $G$ is free abelian of rank $n$ iff it is isomorphic to ${\mathbb{Z}}^{\oplus n}$. We can now call ${\mathbb{Z}}^{\oplus n}$ as the free abelian group of rank $n$.

Proof.

I’m pretty sure that the proofs of Prp 102.6 and Prp 102.7 work verbatim here.□

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The Invariant Factor Theorem

The key theorem is that a subgroup of a free abelian group is free abelian of rank not exceeding that of the bigger group.

Theorem 103.5(Theorem).

If $H$ is a subgroup of a free abelian group $G$ of rank $n$, then $H$ is free abelian of rank $r⩽n$. Further, there exist bases $\{e_1,\dots ,e_n\}$ of $G$ and $\{d_1{e}_1,\dots ,d_r{e}_r\}$ of $H$ respectively where $d_i$ divides $d_{i+1}$ for $i<r$. The integers $d_i$ are uniquely determined up to sign and are called the invariant factors of $H$.

Proof.

The proof is carried out by induction on $n$ using the division algorithm as follows. It is clear for $n=1$. Assume $n>1$ and that the theorem holds for $m<n$. Corresponding to any basis for $G$, there is a positive integer with the property that it is the smallest positive integer that occurs as a coefficient in the expression of elements of $H$ in terms of this basis. Let $l_1$ be the smallest such integer with respect to all bases of $G$. Let $v_1,\dots ,v_n$ be a corresponding basis for $G$ and $v\in H$ be the element such that

$$v=l_1{v}_1+\sum _{i=2}^n{a}_i{v}_i\in H.$$

Writing $a_i=q_i{l}_1+r_i$ with $0⩽r_i<l_1$, we get

$$v=l_1\left(v_1+\sum _{i=2}^n{q}_i{v}_i\right)+\sum _{i=2}^n{r}_i{v}_i.$$

Let $w_1:=v_1+{\sum }_{i=2}^n{q}_i{v}_i$. Observe that $w_1,v_2,v_3,\dots ,v_n$ is a basis of $G$. By the minimality of $l_1$, we must have $r_i=0$ for all $2⩽i⩽n$, and we have $v=l_1{w}_1$.

Consider the subset $H_0$ of $H$ whose elements have coefficient of $w_1$ to be zero when written in terms of the basis $w_1,v_2,v_3,\dots ,v_n$. Clearly, $H_0$ is a subgroup of $H$ and $H_0\cap \mathbb{Z}v=\{0\}$. Also, if $h\in H$, we can write $h=b_1{w}_1+{\sum }_{i=2}^n{b}_i{v}_i$. If $b_1=m_1{l}_1+s_1$ with $0⩽s_1<l_1$, we have

$$\begin{array}{r}h-m_{1}v=s_1{w}_1+\sum _{i=2}^n{b}_i{v}_i.\end{array}$$

Since $h-m_{1}v\in H$, $s_1=0$ by the minimality of $l_1$. Thus, every $h\in H$ can be expressed in the from

$$h=\underset{\in \mathbb{Z}v}{\underbrace{m_{1}v}}+\underset{\in H_0}{\underbrace{\sum _{i=2}^n{b}_i{v}_i}}.$$

It follows that $H=\mathbb{Z}v\oplus H_0$.

Now, $H_0$ is contained in the subgroup $G_0={⨁}_{i=2}^n\mathbb{Z}{v}_i$. By the induction hypothesis, $G_0$ has a basis $w_2,\dots ,w_n$ and there exists $r⩽n$ such that $H_0$ has a basis of the form $d_2{w}_2,\dots ,d_r{w}_r$ with $d_2|d_3|\dots |d_r$. Clearly, therefore, $H$ itself has rank $r$ and $l_1{w}_1,d_2{w}_2,\dots ,d_r{w}_r$ is a basis for $H$. We only have to show that $l_1|d_2$.

Once again, write $d_2=c{l}_1+d$ with $0⩽d<l_1$ and notice that $l_1{w}_1+d_2{w}_2=l_1(w_1+c{w}_2)+d{w}_2\in H$. Since $w_1+c{w}_2,w_2,w_3,\dots ,w_n$ is a basis for $G$, $d$ is forced to be zero by the minimality of $l_1$, i.e, $l_1|d_2$.□

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Corollary 103.6(Structure theorem for finitely generated abelian groups).

A finitely generated abelian group is isomorphic to ${\mathbb{Z}}^m\times {\mathbb{Z}}_{d_1}\times \cdots \times {\mathbb{Z}}_{d_r}$ for some $m⩾0$ and $d_i$ dividing $d_{i+1}$. The integer $m$ as well as all the $d_i$‘s (up to sign) are uniquely determined.

Proof.

Let $G$ be a finitely generated abelian group with generating set $X$ with $|X|=n$. Let $F$ be the free abelian group on $X$. By Def 1, we have a map $\psi :F\twoheadrightarrow G$ (which is surjective since $X$ generates $G$). Thus, we can write $G\cong F/\ker \psi$. Changing bases to the one prescribed by Thm 5 and using Prp 95.8, we have

$$\begin{array}{r}G\cong \frac{{\mathbb{Z}}^{\oplus n}}{d_1\mathbb{Z}\oplus d_2\mathbb{Z}\oplus \cdots \oplus d_r\mathbb{Z}}\cong \frac{\mathbb{Z}}{d_1\mathbb{Z}}\oplus \cdots \oplus \frac{\mathbb{Z}}{d_r\mathbb{Z}}\oplus {\mathbb{Z}}^{\oplus (n-r)}.\end{array}$$ □