1.2 Free Abelian Group

In this section, we write abelian group additively.

Definition

Let $F$ be an abelian group. For a subset $X\subseteq F$, define $⟨X⟩=\{n_1{x}_1+\cdots +n_t{x}_t:n_i\in \mathbb{Z},x_i\in X\}$. We say $X$ is a basis if

• $⟨X⟩=F$, and
• if ${\sum }_{i=1}^t{n}_i{x}_i=0$ for some $n_i\in \mathbb{Z}$ and $x_i\in X$, then $n_i=0$ for all $i$.

We say $F$ is a free abelian group if it has a basis.

Definition

Let $\{G_i{\}}_{i\in I}$ be a family of groups. Define

• (external) direct product as ${\Pi }_{i\in I}{G}_i:=\{(a_i{)}_{i\in I}:a_i\in G_i\}$.
• (external) weak direct product as ${\Pi }_{i\in I}^{\prime }{G}_i:=\{(a_i{)}_{i\in I}:a_i\in G_i,a_i=1\text{mbox}foralmostalli\}$
• Suppose $G_i$ are additive abelian groups, define (external) direct sum as ${\oplus }_{i\in I}{G}_i:={\Pi }_{i\in I}^{\prime }{G}_i$.

Remark. The definition of (iii) is different from Hungerford.

Definition

Let $\{N_i:i\in I\}$ where $N_i⊲G$. If

• $⟨{\cup }_{i\in I}{N}_i⟩=G$, and
• $\forall k\in I$, $N_k\cap ⟨{\cup }_{i\ne k}{N}_i⟩=\{e_G\}$,

then we say $G$ is the internal weak direct product of $N_i$, or internal direct sum of $N_i$ (if written additively). Conversely, $G\simeq {\Pi }_{i\in I}^{\prime }{N}_i$.

Theorem

Let $F$ be an abelian group. TFAE:

• $F$ has a non-empty basis $X$.
• $F$ is the (internal) direct sum of a family of infinite cyclic groups.
• $F\simeq {\oplus }_{i\in I}\mathbb{Z}$ is an (external) direct sum of copies of $(\mathbb{Z},+)$.

\begin{proof} i) → ii). For any $x\in X$, $nx=0$ iff $n=0$. Thus $⟨x⟩$ is an infinite cyclic subgroup. Claim: $F$ is the internal direct sum of $⟨x⟩$ for all $x\in X$. By definition, it suffices to check that $⟨x⟩\cap ⟨X\setminus \{x\}⟩=\{0\}$. Indeed, if $nx={\sum }_{x_i\ne x}{n}_i{x}_i$, then $n=0$ and $n_i=0$. So we finish the proof.

ii) → iii). Easy.

iii) → i). Easy. \end{proof}

Theorem

Let $F$ be a finite generated abelian group. Then any two bases of $F$ have the same cardinality.

\begin{proof} By ^2408e8, suppose $F\simeq {\oplus }_{i\in I}\mathbb{Z}$ and $F\simeq {\oplus }_{j\in J}\mathbb{Z}$. Then $F/2F\simeq {\oplus }_{i\in I}\mathbb{Z}/2\mathbb{Z}$ and $F/2F\simeq {\oplus }_{j\in J}\mathbb{Z}/2\mathbb{Z}$. Then $|F/2F|=2^{|I|}=2^{|J|}$. Thus $|I|=|J|$. \end{proof}

Definition

The rank of finite generated abelian group $F$ is the cardinality of basis, denoted by $\mathrm{rk}F$ or ${\mathrm{rk}}_{\mathbb{Z}}F$.

Corollary

Let $F_1,F_2$ be two finite generated abelian group. Then $F_1\simeq F_2$ iff $\mathrm{rk}{F}_1=\mathrm{rk}{F}_2$.

Theorem

Let $G$ be an abelian group. Then it is the homomorphic image of a free abelian group of rank $|X|$, where $X$ is a set of generators of $G$.

\begin{proof} Let $F:={\oplus }_{x\in X}\mathbb{Z}$. Define $f:F\to G,(n_{x}x{)}_{x\in X}\mapsto {\sum }_{x\in X}{n}_{x}x$. Then $G=\mathrm{Im}f$. \end{proof}

Theorem

Let $M$ be a finite generated free abelian group with ${\mathrm{rk}}_{\mathbb{Z}}M=n$. Let $M^{\prime }<M$ nonzero. Then

• $M^{\prime }$ is also a free abelian group with $\mathrm{rk}{M}^{\prime }=q⩽n$.
• There exists some basis $(e_1,\cdots ,e_n)$ of $M$ and $a_1,\cdots ,a_q\in {\mathbb{Z}}^{+}$ such that $a_1\mid a_2\mid \cdots \mid a_q$ such that $(a_1{e}_1,\cdots ,a_q{e}_q)$ is a basis of $M^{\prime }$.

Before the proof, we firstly give the following examples.

Examples.

• $M=\mathbb{Z}$ and $M^{\prime }=2\mathbb{Z}$. Then $e_1=1$ and $a_1=2$.
• $M=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $M^{\prime }=⟨2e_1+3e_2⟩$. If we let $f_1=2e_1+3e_2$ and $f_2=e_1+2e_2$, then $(f_1,f_2)$ is a basis of $M$ and $1\cdot f_1$ is a basis of $M^{\prime }$ with $a_1=1$.
• $M=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $M^{\prime }=⟨3e_1,4e_2⟩$. Let $f_1=3e_1+4e_2$ and $f_2=2e_1+3e_2$. Then $M^{\prime }=⟨f_1,12f_2⟩$.

How to compute $a_1,\cdots ,a_q$?

Let $M=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $M^{\prime }=⟨2e_1+3e_2,4e_1+12e_2⟩$. $M^{\prime }$ can be written as $\left[\begin{array}{cc}2& 3\\ 4& 12\end{array}\right]\left[\begin{array}{c}{e}_1\\ e_2\end{array}\right]$. Note that

$$\begin{array}{rl}& \left[\begin{array}{cc}2& 3\\ 4& 12\end{array}\right]\left[\begin{array}{c}{e}_1\\ e_2\end{array}\right]\\ =& \left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& 6\end{array}\right]\left[\begin{array}{c}{e}_1\\ e_2\end{array}\right]\\ =& \left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 6& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}-12& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]\left[\begin{array}{c}{e}_1\\ e_2\end{array}\right]\\ =& M\left[\begin{array}{cc}-12& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}{e}_1+e_2\\ 2e_1+3e_2\end{array}\right].\end{array}$$

where $M$ is a invertible matrix. So $M^{\prime }=⟨f_1,12f_2⟩$ where $f_1=2e_1+3e_2$ and $f_2=e_1+e_2$.

Now we prove ^93020f and we start with a lemma.

Lemma

Let $f:M\to N$ be an epimorphism of abelian groups, and suppose $N$ is a finite generated free abelian group. Then $M\simeq N\oplus \ker f$ as abelian groups.

\begin{proof} Let $\{e_1,\cdots ,e_n\}$ be a basis of $N$. Suppose $x_i\in M$ such that $f(x_i)=e_i$. Define a map $g:M\to N\oplus \ker f$ where for $x\in M$, suppose $f(x)={\sum }_{i=1}^n{a}_i{e}_i$, $a_i\in \mathbb{Z}$, then let $g(x)=(f(x),x-{\sum }_{i=1}^n{a}_i{x}_i)\in N\oplus \ker f$. Then it suffices to check $g$ is an isomorphism.

• Homomorphism: It is easy to verify $g$ is a homomorphism.
• Injective: If $f(x)=x-{\sum }_{i=1}^n{a}_i{x}_i=0$, then $\sum a_i{e}_i=0$, $a_i\equiv 0$ and so $x=0$. Thus $g$ is injective.
• Surjective: Given $(\alpha ,\beta )\in N\oplus \ker f$, then $\alpha ={\sum }_{i=1}^n{a}_i{e}_i$. Let $x={\sum }_{i=1}^n{a}_i{x}_i+\beta \in M$. Then $g(x)=(\alpha ,\beta )$. \end{proof}

Remark. $0\to \ker f\to M\stackrel{f}{\to }N\to 0$ is a short exact sequence. If $N$ is free, then there exists $s:N\to M$ such that $f\circ s={\mathrm{id}}_N$.

\begin{proof} It is a proof of ^93020f, (i).

(i) By induction on $n$. When $n=1$, then $M=\mathbb{Z}e>M^{\prime }\ne 0$. Elements in $M$ are of form $ae$ for some $a\in \mathbb{Z}$. Let $a_1=\min \{a\in \mathbb{Z}:ae\in M^{\prime }\}$. Claim that $M^{\prime }=⟨a_{1}e⟩$. Indeed, $⟨a_{1}e⟩\subseteq M^{\prime }$. If it is not equal, then there exists $ae\in M^{\prime }$ and $ae\notin ⟨a_{1}e⟩$. WLOG $a>0$. Since $ae\notin ⟨a_{1}e⟩$, we have that $a_1\nmid a$ and $(a_1,a)=b<a_1$. Then by Bezout’s theorem there exists $x,y\in \mathbb{Z}$ such that $ax+a_{1}y=b$, and so $x(ae)+y(a_{1}e)=be\in M^{\prime }$. But $0<b<a_1$, which contradicts with minimality of $a_1$.

Now suppose (i) holds when $\mathrm{rk}M⩽n$. Now consider $\pi :M=\mathbb{Z}\oplus {\mathbb{Z}}^n\twoheadrightarrow {\mathbb{Z}}^n,(a,b)\mapsto b$. Let $N=\pi (M^{\prime })<{\mathbb{Z}}^n$. By induction hypothesis, $N$ is free with $\mathrm{rk}N⩽n$. Also we know $\pi$ induces a map $\pi {|}_{M^{\prime }}:M^{\prime }\twoheadrightarrow N$. By ^f38221, $M^{\prime }\simeq N\oplus \ker (\pi {|}_M^{\prime })$. Note that $\ker \pi =\mathbb{Z}\oplus 0$. Thus $\ker (\pi {|}_{M^{\prime }})<\mathbb{Z}\oplus 0$ is free, whose rank is less equal than $1$. Therefore, $M^{\prime }$ is free and $\mathrm{rk}⩽n+1$.

(ii) When $n=1$, the statement is trivial (see (i)). Consider the case of $n=2$. By ^f38221, $M^{\prime }$ can be decomposed a s $M^{\prime }=\pi (M^{\prime })\oplus \ker \pi {|}_{M^{\prime }}$ where $\pi (M^{\prime })⩽\mathbb{Z}$ and $\ker \pi {|}_{M^{\prime }}⩽\mathbb{Z}\oplus 0$. By case of $n=1$, we get $M^{\prime }=a_1{f}_1^{\prime }\oplus a_2{f}_2^{\prime }$ for some basis $f_1^{\prime },f_2^{\prime }$ and $a_1,a_2⩾0$. If one of $a_i=0$, we have done. So suppose $a_i\ne 0$, $i=1,2$. (naive idea: hope $a_1\mid a_2$, i.e. want $a_1$ to be “small”.) Consider the set

$$P=\{a\in {\mathbb{Z}}_{⩾0}:\exists b\in {\mathbb{Z}}_{⩾0},\text{mbox}andthereexistsabasis(f_1^{\prime \prime },f_2^{\prime \prime })\text{mbox}ofM\text{mbox}suchthat{M}^{\prime }=⟨a{f}_1^{\prime \prime },b{f}_2^{\prime \prime }⟩\}.$$

Since $a_1\in P$ and $P$ is non-empty, then there exists unique element $\alpha \in P$. So $M^{\prime }=⟨\alpha f_1,\beta f_2⟩$. Now claim that $\alpha \mid \beta$. Otherwise, $(\alpha ,\beta )=t<\alpha$. Write $\alpha =t{x}_1$ and $\beta =t{x}_2$ and let $t=\alpha y_1+\beta y_2$. Then $x_1{y}_1+x_2{y}_2=1$. Note that $M^{\prime }=⟨\alpha f_1,\beta f_2⟩=⟨t(x_1{f}_1+x_2{f}_2),t{x}_1{x}_2(-y_2{f}_1+y_1{f}_2)⟩$ by $\det \left[\begin{array}{cc}1& 1\\ -x_2{y}_2& x_1{y}_1\end{array}\right]=1$. In addition, $(x_1{f}_1+x_2{f}_2,-y_2{f}_1+y_1{f}_2)=(f_1,f_2)\left[\begin{array}{cc}{x}_1& -y_2\\ x_2& y_1\end{array}\right]$ and $\det \left[\begin{array}{cc}{x}_1& -y_2\\ x_2& y_1\end{array}\right]=1$ yield that $(x_1{f}_1+x_2{f}_2,-y_2{f}_1+y_1{f}_2)$ is a basis. Thus $t\in P$, which contradicts with minimality of $\alpha$. So $\alpha \mid \beta$. Now we finish the proof of the case $n=2$. (For example, let $M^{\prime }=⟨10e_1,15e_2⟩<⟨e_1,e_2⟩$. Use above algorithm, $(\alpha ,\beta )=(10,15)=5=t$. Then $x_1=2,x_2=3$ and $y_1=-1,y_2=1$. It yields that $M^{\prime }=⟨5(2e_1+3e_2),30(-e_1-e_2)⟩$. )

Now suppose that the statement is true for $n$-case. now suppose $M={\mathbb{Z}}^{n+1}$. Again, $M^{\prime }=\ker {|}_{M^{\prime }}\oplus \pi (M^{\prime })$ where $\pi (M)⩽{\mathbb{Z}}^n$ and $\ker {|}_{M^{\prime }}⩽\mathbb{Z}\oplus 0$. By induction hypothesis, $\pi (M^{\prime })=⟨a_1{f}_1,a_2{f}_2,\cdots ,a_{n+1}{f}_{n+1}⟩$ where $a_2\mid a_3\mid \cdots \mid a_{n+1}$. Let

$$P=\{a\in {\mathbb{Z}}_{+}:\exists b_2\mid b_3\mid \cdots \mid b_{n+1}\text{mbox}and\exists (f_1,\cdots ,f_{n+1})\text{mbox}suchthat{M}^{\prime }=⟨a_1{f}_1,\cdots ,a_{n+1}{f}_{n+1}⟩\}.$$

Since $a_1\in P$, $P$ is non-empty. Then let $\alpha$ be the minimal element of $P$. Claim that $\alpha \mid b_2$. We start with checking $\alpha \mid {\beta }_{n+1}$. Otherwise, $(\alpha ,{\beta }_{n+1})=t<\alpha$ and we can use the trick in the case of $n=2$. Note that $M^{\prime }=⟨t(x_1{f}_1+x_2{f}_{n+1},{\beta }_2{f}_2,\cdots ,{\beta }_n{f}_n,t{x}_1{x}_2(-y_2{f}_1+y{f}_{n+1}))⟩$ and so $t\in P$, contradiction. Next, we check $\alpha \mid {\beta }_n$. If not, then similar to above we can change the sequence $\alpha ,{\beta }_2,\cdots ,{\beta }_{n-1},{\beta }_n,{\beta }_{n+1}$ to $t,{\beta }_2,\cdots ,{\beta }_{n-1},t{x}_1{x}_2,{\beta }_{n+1}$, where $\alpha =t{x}_1$ and ${\beta }_n=t{x}_2$. Claim that $t{x}_1{x}_2\mid {\beta }_{n+1}$. Indeed, since $t{x}_1\mid {\beta }_{n+1}$ and $t{x}_2\mid {\beta }_{n+1}$, we have $[t{x}_1,t{x}_2]\mid {\beta }_{n+1}$ with $[t{x}_1,t{x}_2]=t{x}_1{x}_2$. Therefore, $t\in P$, which is impossible. So $\alpha \mid {\beta }_n$. Repeat the procedure, we can prove that $\alpha \mid {\beta }_1$ and so we finish the proof. \end{proof}

Corollary

Let $G$ be a finitely generated abelian group, which is generated by $n$ elements. Then any $H<G$ can be generated by $m$ elements with $m⩽n$.

\begin{proof} By ^d4cf25, there exists free abelian group $F$ of rank $n$ giving rise to an epimorphism $\pi :F\to G$. Since ${\pi }^{-1}(H)<F$, by ^93020f ${\pi }^{-1}(H)$ is free, whose rank is less equal than $n$. As ${\pi }^{-1}(H)\twoheadrightarrow H$, the image of a basis of ${\pi }^{-1}(H)$ gives a set of generators of $H$. \end{proof}