# Ranks, Tensors and Abelian Groups

2016-11-09 ( 2016-11-09)## Abelian Groups

Given $a\in A$, the homomirphism
$i_ a \from \ZZ \to A$ defined by
$i_ a(k) = ka$
is mono (injective) iff $a$ has infinite order.
Of $S$ is a set,
the **free** Abelian group generated by $S$
is
$$
\ZZ[S] = \ZZ S = \bigoplus_ {s \in S} \ZZ,
$$
and it is the group of all finite linear combinations (with integer coefficients)
of elements in $S$. The elements of $S$ are a **basis** (**base**) of $\ZZ[S]$.
If $A$ is an Abelian group, and $S\subset A$ is a set of elements such that
the homomorphism
$$\bigoplus_ {s\in S} i_ s\from \bigoplus_ {s \in S} \ZZ \to A$$
is an isomorphism,
then $A$ is a **free** group generated by the elements
in $S$ (its **basis**).
More on abelian groups in the monograph of Fuchs (2015)^{1}.

**Universal property:**
If $S\subset A$ is a basis, $B$ is any Abelian group,
for every function $\varphi \from S \to B$
there exists a **unique** group homomorphism $f\from A \to B$ such that
$f(s) = \varphi(s),$ for all $s\in S.$
In other words, homomorphisms from a free group are completely and
arbitrarily determined by their values on the elements of a basis.

- (a)
- Every subgroup of a free abelian group is free.

*Proof:*
This theorem is credited to Richard Dedekind in Johnson - 1980 (EBSCO LINK), page 8.
For a nice introduction to Abelian groups, and the proof of **(a)** which I took as source,
see Kurosh^{2}, §19.
A more general approach:
Lang^{3}, Chapter 1, §7 and page 41.
We give here a proof for a countable basis.
Let $a_ 1, a_ 2, \ldots, a_ n, \ldots$ be elements of a basis $S$ of a free group $F$.
Hence $F$ is isomorphic to the group of all functions $S\to \ZZ$ with finite support.
With an abuse of notation, assume $F$ is the group of such functions.
$\newcommand{\supp}{\operatorname{supp}}$
For each $f\in F$, $\supp(f) = \{ s \in S : f(s) \neq 0 \}$
denotes the support of $f$.
Being finite, for each $f\in F$, the maximum
$\max \supp(f)$ exists whenever $f\neq 0$.
It is the *last index* of $f$. The corresponding
coefficient $f(s)$ is the *last coefficient* of $f$.

Let $A\subset F$ be a fixed subgroup. Let $b_ 1$ be the non-zero element in $A$ with smallest last index, say $s_ 1$, and smallest positive last coefficient. Then, $b_ 1 \in A$, and if $x\in A$, $x\neq 0$,then either $\max \supp(x) > \max \supp(b_ 1)=s_ 1$ or $\max \supp(x) = \max \supp(b_ 1 = s_ 1)$. Let $A_ 1 = \{ a \in A : \max \supp(a) = s_ 1 \} $. If $a\in A_ 1$, then by definition $\max \supp(a) = s_ 1$, then $|a(s_ 1)| \geq b_ 1(s_ 1)$, and hence there are integers $q$ and $r$ such that $0\leq r < b_ 1(s_ 1)$ and $|a(s_ 1)| = q b_ 1(s_ 1) + r $. But if $r\neq 0$ then $q b_ 1 - a$ has last coefficient equal to $r$, which is smaller than $b_ 1(s_ 1)$, so it must be $r=0$. It therefore exists $q$ such that $a(s_ 1) = q b_ 1(s_ 1)$. Now, if $q b_ 1 - a$ were not $0$ in $A$, its last index would be smaller than $s_ 1$, which would be a contradiction, and hence there exists $q\in \ZZ$ such that $a= q b_ 1$. In other words, $A_ 1 = \langle b_ 1 \rangle = \ZZ[b_ 1]$.

For the next step, define as $b_ 2$, the element in $A \smallsetminus A_ 1$ with smallest last index (say, $s_ 2$) and smallest nonzero positive last coefficient. Since $b_ 2 \not\in A_ 1$, $s_ 1 < s_ 2$. Let $A_ 2 = \{ a \in A : \max \supp(a) \leq s_ 2 \} $. Note that $A_ 1 \cap \langle b_ 2 \rangle = 0$, and therefore it is a direct sum $A_ 1 \oplus \langle b_ 2 \rangle \subset A_ 2$. Moreover, if $a \in A_ 2\smallsetminus A_ 1$, then $\max \supp (a) = s_ 2$, and as before there exist $q \in \ZZ$ such that $a (s_ 2) = q b_ 2(s_ 2)$ (otherwise $b_ 2(s_ 2)$ is not the smallest positive last coefficient), which means that $q b_ 2 - a \in A_ 1=\langle b_ 1\rangle $, or that $a \in \langle b_ 1 \rangle \oplus \langle b_ 2 \rangle$. Thus, $A_ 2 = \ZZ[b_ 2] \oplus A_ 1$.

By induction, we can continue with a sequence of subgroups
$$
A_ 1 \subset A_ 2 \subset \ldots \subset A_ n \subset \ldots \subset A
$$
such that $A_ n$ is the free group generated by the elements
$b_ 1, \ldots, b_ n, \ldots$ (which are a basis for $A_ n$).
This sequence either stops, when $A_ n = A$,
and hence $A$ is a free group of rank $n$,
or it never stops,
and $A$ is the free group generated by the set of all $b_ j$
(each element $a \in A$ has a last index;
there must be an $s_ j$ such that $s_ j$ is greater than this
last index, and therefore $a\in A_ j$).
*/qed/*

- (b)
- Every subgroup of a finitely generated abelian group is finitely generate.

*Proof:*
See Kurosh, §20, or Fuchs Theorem 2.5 pag. 82, or Theorem 2.6 pag. 83 (Stacked Basis Theorem).
*/qed/*

From the previous propositions and the Smith Normal Form of a $\ZZ$-matrix, one can (constructively) prove that

- (c)
- A finitely generated Abelian group is the finite direct sum of cyclic subgroups $C_ j\subset A$,
$$
A \cong \bigoplus_ {j=1}^l C_ j
$$
with $C_ j$ cyclic of order $\nu_ j$ (or order infinity if $\nu_ j= 0$,
i.e. $C_ j = \ZZ / \nu_ j \ZZ$, with $\nu_ j \geq 0$.

The terms with $\nu_ j\geq 1$ contribute to the **torsion** subgroup of $A$,
while the terms with $\nu_ j = 0$ contribute to the **torsion-free** rank
of $A$. In other words,
$A= T \oplus \ZZ^r$, where $r$ is the **rank** of $A$. We will show some properties and
alternative definitions of the rank (for example, if $A$ is not finitely generated).

$\require{AMScd}$

## Tensor product of $\ZZ$-modules (AKA Abelian Groups)

Let $A$, $B$ be two Abelian groups.
The **tensor** product $A\otimes_ \ZZ B = A \otimes B $ is defined as follows:
let $\ZZ[A\times B]$
denote the free abelian group generated by elements in the cartesian product
$A\times B$ (i.e., finite $\ZZ$-linear combinations of symbols in $A\times B$).
Elements of $A\times B$ will be written as $a\times b$ instead of $(a,b)$.
Then, consider in $\ZZ[A\times B]$ the subgroup $R$ generated by
$$
\begin{aligned}
(a_ 1 + a_ 2) \otimes b - (a_ 1 \otimes b + a_ 2 \otimes b) & \text{ with } a_ 1, a_ 2\in A, b \in B \\

a \otimes (b_ 1 + b_ 2) - (a \otimes b_ 1 + a \otimes b_ 2) & \text{ with } a\in A, b_ 1, b_ 2 \in B \\

a k \otimes b - a \otimes k b & \text{ with } a\in A, b\in B, k \in \ZZ

\end{aligned}
$$
*(note: the third line is not independent, actually, for $\ZZ$-modules)*
The tensor product is the quotient
$$
A \otimes B = \dfrac{ \ZZ[A\times B] }{R}~.
$$
The coset containing the element $(a,b) \sim a \otimes b$
will be denoted again as $a \otimes b$.
It is an Abelian Group (a $\ZZ$-module), as quotient of a free abelian group and $R$.

The tensor product has the followin *Universal Property*:
a function $f \from A \times B \to G$ is *bilinear* if
$$
\begin{aligned}
f (a_ 1 + a_ 2, b ) =f (a_ 1, b) + f(a_ 2, b) & \text{ with } a_ 1, a_ 2\in A, b \in B \\

f( a , b_ 1 + b_ 2) = f( a , b_ 1) + f ( a , b_ 2) & \text{ with } a\in A, b_ 1, b_ 2 \in B \\

f( a k , b) = f( a , k b) & \text{ with } a\in A, b\in B, k \in \ZZ

\end{aligned}
$$
*(note: the third line is not independent, actually, for $\ZZ$-modules)*
For evary abelian group $G$ and every bilinear map $f\from A\times B \to G$,
there exists a **unique** homomorphism of abelian groups
$F\from A\otimes B \to G$ such that
$F(a\times b) = f(a,b)$.

- (1)
- $A\otimes B \cong B \otimes A$.
- (2)
- $\ZZ \otimes A \cong A \otimes \ZZ \cong A $.

*Proof:*
The isomorphism is given by $k\otimes a \in \ZZ \otimes A \mapsto ka \in A$,
with inverse
$a \mapsto 1\otimes a$. */qed/*

- (3)
- If $\alpha \from A \to A'$ is a homomorphism,
then $\alpha_ *\from A \otimes B \to A' \otimes B$
defined by $\alpha_ *(a\otimes b) = (\alpha a) \otimes b$,
for $a\in A$, $b\in B$, is a homomorphism.
The same for $\beta\from B \to B'$,
$\beta_ *\from A\otimes B \to A \otimes B'$ is the homomorphism defined as
$\beta_ *(a\otimes b) = a \otimes (\beta b)$.
Both $-\otimes B$ and $A \otimes -$ are covariant functors from
the category of Abelian groups to the category of Abelian groups.

Given two morphisms $\alpha \from A \to A'$ and $\beta \from B \to B'$, the notation $\alpha \otimes \beta \from A\otimes B \to A' \otimes B'$ denotes the composition $\alpha_ * \beta_ *$.

- (4)
- The functor $A \otimes -$ is additive: there is a
*natural isomorphism*$$ A \otimes

\left( \bigoplus_ {j\in J} B_ j \right) \cong A \otimes \bigoplus_ {j\in J} B_ j $$

*Proof:*
Cfr. Proposition 7.3, page 110 of Hilton-Stammbach (1997)^{4}
*/qed/*

- (5)
- The functor $A\otimes -$ is right-exact:
if $B'\stackrel{\beta'}{\to} B \stackrel{\beta''}{\to} B'' \to 0$
is an exact sequence of abelian groups,
then for any abelian group $A$ the sequence
\begin{CD}
A \otimes B' @>{\beta'_ *}>> A \otimes B @>{\beta''_ *}>> A \otimes B'' @>{}>> 0
\end{CD}

is exact.

*Proof:*
Cfr. Proposition 7.3, page 110 of Hilton-Stammbach (1997) */qed/*

An abelian group (a $\ZZ$-module, and more generally an $R$-module) $A$ is called **flat**
if the corresponding tensor functor $A \otimes -$ is not only right-exact, but it is exact:
more precisely, for every short exact sequence
$0 \to B'\stackrel{\beta'}{\to} B \stackrel{\beta''}{\to} B'' \to 0$
the sequence
\begin{CD}
0 @>{}>> A \otimes B' @>{\beta'_ *}>> A \otimes B @>{\beta''_ *}>> A \otimes B'' @>{}>> 0
\end{CD}

is exact.

- (6)
- An abelian group $A$ is flat if and only if it is torsion-free. Hence $\QQ \otimes -$ is an exact functor.

*Proof:*
Exercise 8.7 of Hilton-Stammbach, page 115.
Note that torsion-free $\neq$ free (e.g. $\QQ$).
*/qed/*

## Rank of an abelian group

Let $A$ be an abelian group. Its **rank** is the maximal number of
$\ZZ$-*linearly independent* elements in $A$.
Recall that $n$ elements $a_ 1, a_ 2, \ldots, a_ n$ in
$A$ are dependent if there are $n$ integers
$k_ 1, k_ 2, \ldots, k_ n$, not all zero,
such that
$k_ 1 a_ 1 + k_ 2 a_ 2 + \ldots + k_ n a_ n = 0$.
And they are ($\ZZ$-linearly) independent if they
are not dependent.
Each element of a set of independent elements in $A$ has infinite order.
A group $A$ is **free** if it is generated by a set of independent elements
(i.e., a **basis**).

**Remark:** This definition is actually the definition of $0$-rank,
or torsion-free rank, for some authors (Robinson^{5}, page 96).

**Example:**
$\QQ$ has rank $1$, and it is not finitely generated.

**Example:**
$\ZZ^r$ has rank $r$.

**Example:** $\QQ/\ZZ$ is equal to its torsion subgroup (see below), and has rank $0$ (and it is not finitely generated).

- (7)
- The rank of an Abelian group $A$ is the $\QQ$-dimension of the $\QQ$-vector space $\QQ \otimes A$: $$ \operatorname{rk} A = \dim_\QQ \QQ \otimes A $$

*Proof:*
See the proof of 4.2.1, page 96, of Robinson.
*/qed/*

Let $A,B,C$ be abelian groups,
and $f\from A \to B$ a monomorphism (injective homomorphism)
$g\from B \to C$ an epimorphism (surjective homomophism)
such that $\operatorname{Im} f = \ker g$. In other words, the following
sequence is exact.
\begin{CD}
0 @>{}>> A @>{f}>> B @>{g}>> C @>{}>> 0 \\

\end{CD}

- (8)
- If the previous sequence is exact, then $\operatorname{rk} B = \operatorname{rk}{A} + \operatorname{rk}{C}$

*Proof:*
Since the functor $\QQ \otimes -$ is exact, as $\QQ$ is flat (torsion-free),
the sequence
\begin{CD}
0 @>{}>> \QQ \otimes A @>{f_ *}>> \QQ \otimes B @>{g_ *}>> \QQ \otimes C @>{}>> 0 \\

\end{CD}
is exact (and it is a sequence of $\QQ$-vector spaces).
Therefore
$\dim \QQ \otimes B = \dim \QQ \otimes A + \QQ \otimes C$,
$\implies$
$\operatorname{rk} B = \operatorname{rk}{A} + \operatorname{rk}{C}$
*/qed/*

- (9)
- Let $T\subset A$ the torsion subgroup of $A$, consisting of all elements with finite order in $A$. The quotient $A/T$ is torsion-free, and $\operatorname{rk}{A} = \operatorname{rk}{A/T}$.

*Proof:*
It is easy to see that $\operatorname{rk}{T} = 0$,
and since the sequence $0 \to T \to A \to A/T \to 0 $ is exact,
$\operatorname{rk} A = 0 + \operatorname{A/T}$.
*/qed/*

- (A)
- The rank of a free Abelian group is the number of elements in any basis. The rank of an arbitrary Abelian group $A$ is the supremum of the rank of $F$, where $F$ ranges over all the free subgroups $F\subset A$.

*Proof:*
A free abelian group is isomorphic to $\ZZ^r$ (if finitely generated) or $\ZZ^\infty$ (if not finitely generated),
and the first part follows easily.
Note that if $F\subset A$, then
$\operatorname{rk} A = \operatorname{rk} F +
\operatorname{rk} A/F$, and hence
$\operatorname{rk}{F} \leq \operatorname{rk} A$.

Now, consider the torsion subgroup $T\subset A$. The quotient $T/A$ is torsion-free, and
the ranks of the free subgroups of $A$ are the same as the ranks of the free subgroups of $T/A$ (in fact,
the families of free subgroups are almost the same : why?).
Now the result follows by the same fact on the dimension of $\QQ$-vector spaces.
*/qed/*

- Fuchs, László
*Abelian groups.*Springer Monographs in Mathematics. Springer, Cham, 2015. xxi+747 pp. ISBN: 978-3-319-19421-9; 978-3-319-19422-6 20Kxx (20-02)(http://link.springer.com/book/10.1007%2F978-3-319-19422-6)^{[return]} - Kurosh, A. G.
*The theory of groups. Vol. I.*Translated and edited by K. A. Hirsch. Chelsea Publishing Co., New York, N.Y., 1955. 272 pp. 20.0X LINK^{[return]} - Lang, Serge
*Algebra.*Revised third edition. Graduate Texts in Mathematics, 211. Springer-Verlag, New York, 2002. xvi+914 pp. ISBN: 0-387-95385-X^{[return]} - Hilton, P. J.; Stammbach, U.
*A course in homological algebra.*Second edition. Graduate Texts in Mathematics, 4. Springer-Verlag, New York, 1997. xii+364 pp. ISBN: 0-387-94823-6^{[return]} - Robinson, Derek J. S.
*A course in the theory of groups.*Second edition. Graduate Texts in Mathematics, 80. Springer-Verlag, New York, 1996. xviii+499 pp. ISBN: 0-387-94461-3^{[return]}