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\begin{document}
\centerline{Math 372 lecture for Monday, Week 11}
\bigskip

\centerline{\bf Finitely generated abelian groups}
\bigskip

An {\em abelian group} is a pair~$(A,+)$ consisting of a set~$A$
and an operation~$+\colon A\times A\to A$ called {\em addition}
such that~$+$ is associative and commutative, there exists $0\in A$ such that~$a+0=0+a=a$
for all~$a\in A$, and each~$a\in A$ has an additive inverse~$-a$ such
that~$a+(-a)=0$.  ({\em Subtraction} is defined using additive inverses:
  $a-b:=a+(-b)$.)  It is {\em finitely generated} if there exists
  and~$a_1,\dots,a_m\in A$  for some~$m$ such that for each~$a\in A$, there
  exists~$n_1,\dots,n_m\in\Z$ such that
  \[
    a =n_1a_1+\dots+n_ma_m.
  \]

  {\bf Examples.}
\begin{itemize}[leftmargin=*]
  \item A {\em cyclic group} is by definition generated by a single element, and every
    cyclic group is abelian.  Every cyclic group is isomorphic
    to~$\Z/n\Z$ for some nonnegative integer~$n$.  The case $n=0$ yields the
    infinite cyclic group $\Z=\Z/0\Z$.
  \item Every finite abelian group is generated by the finite set consisting of all of its
    elements.  For instance, $\Z/4\Z=\langle 0,1,2,3\rangle$.  Of course, we also
    have $\Z/4\Z=\langle 1\rangle=\langle 3\rangle$.  We do not require the set of
    generators to be minimal (with respect to inclusion).
  \item  A finite product of cyclic groups is a finitely generated abelian group.  A typical
    instance is $\Z/4\Z\times\Z/6\Z\times\Z^2$.  Recall that addition is defined
    component-wise, e.g., $(2,5,8,-2)+(3,3,7,2)=(1,2,15,0)$ in this group.
\end{itemize}

The last example is about as complicated as a f.g.~abelian group can get:
it turns out that every f.g.~abelian group is a product of a finite number
of cyclic groups, i.e., has the form~$\prod_{i=1}^k\Z/n_i\Z$ for
suitable~$n_i$ (and where~$n_i$ might be~$0$). The {\em structure} of the group 
is then
given by the list of the~$n_i$. Different choices for the~$n_i$ may produce
isomorphic
groups, but it turns out that the ambiguity is accounted for by the following
well-known result:
\medskip

{\bf Theorem.} (Chinese remainder theorem.) Let~$m,n\in\Z$. Then
\[
\Z/mn\Z\simeq\Z/m\Z\times\Z/n\Z
\]
if and only if~$m$ and~$n$ are relatively prime. If~$\gcd(m, n)=1$, then an
isomorphism is provided by $a\mapsto  (a \bmod m, a \bmod n)$.
\medskip

Thus,~$\Z/24\Z\simeq\Z/8\Z\times\Z/3\Z$, but~$\Z/4\Z\not\simeq\Z/2\Z\times\Z/2\Z$.
\medskip

{\bf Theorem.} (Structure theorem for f.g.~abelian groups)  A group is a
finitely generated abelian group if and only if it is isomorphic to
  \[
    \Z/{n_1}\Z\times\dots\times\Z/{n_k}\Z\times\Z^r
  \]
  for some list (possibly empty) of integers $n_1,\dots,n_k$ with  $n_i>1$ for
  all~$i$ and some integer $r\geq0$. These integers may be required to satisfy
  either of the following two conditions, and in either case they are uniquely
  determined by the isomorphism class of the group.
  
  \noindent{\bf Condition 1:} $n_i|n_{i+1}$ (\,$n_i$ evenly divides $n_{i+1}$) for
  all $i$.  In this case, the~$n_i$ are the {\em invariant factors} of the
  group.

  \noindent{\bf Condition 2:}  There exist primes $p_1\leq\dots\leq p_k$ and
  positive integers $m_i$ such that $n_i=p_i^{m_i}$ for all $i$.  
  In this case, the~$n_i$ are the {\em elementary divisors} and the $\Z/{n_i}\Z$
  are the {\em primary factors} of the group.  
  
  The number $r$ is the {\em rank} of the group.
  \medskip

How does one go about computing the rank and invariant factors of a finitely
generated abelian group $A$?  Let $\{a_1,\dots,a_m\}$ be generators, and define
the group homomorphism determined by
\begin{align*}
  \Z^m&\stackrel{\pi}{\longrightarrow} A\\
  e_i&\mapsto a_i
\end{align*}
where $e_i$ is the $i$-th standard basis vector.  Saying that the $a_i$
generate~$A$ is the same as saying that $\pi$ is surjective.  Next, by a
standard theorem from algebra\footnote{The key point is that abelian groups are
  modules over $\Z$, and $\Z$ is a Noetherian ring.}  every subgroup of $\Z^m$
  is finitely generated.  In particular, there exists a finite set of generators
  $\{b_1,\dots,b_n\}$ for the kernel of $\pi$.  Define  
\[
  \Z^n\stackrel{M}{\longrightarrow}\Z^m
\]
where $M$ is the $m\times n$ integer matrix with $i$-th column $b_i$.  
Combining these two mappings yields a {\em presentation} of $A$:
\[
\begin{tikzcd}
  \Z^n \ar[r,"M"]&\Z^m\ar[r,twoheadrightarrow] &A
\end{tikzcd}
\]
Hence, $\pi$ induces an isomorphism
\begin{align*}
\cok(M):=\Z^m/\textrm{im}(M)&\simeq A\\
e_i\quad&\mapsto a_i,
\end{align*}
where $\cok(M)$ denotes the cokernel of $M\colon\Z^n\to\Z^m$.
In this way, $A$ is determined by the single matrix~$M$.

We have just seen that each finitely generated abelian group is the cokernel of
an integer matrix.  Conversely, each integer matrix determines a finitely
generated abelian group.  However, the correspondence is not bijective.  The
construction of~$M$, above, depended on arbitrary choices for generators of $A$
and of the kernel of $\pi$.  Making different choices creates a different
matrix representing~$A$.  This is especially obvious if we choose a different
number of generators for~$A$ or $\ker{\pi}$.  However, there can be a difference
even if the number of generators is kept constant.  In that case, changing the
choice of generators corresponds to integer changes of coordinates for the
codomain and domain of $M$, or equivalently, to performing integer row and
column operations on $M$.   

Write $M\sim N$ for integer matrices $M$ and $N$ if one may be obtained from the
other through a sequence of integer row and column operations.  Since the
operations are reversible, $\sim$ is an equivalence relation. 

Suppose $M$ is an $m\times n$ integer matrix and $M\sim N$. Start with
identity matrices $P=I_m$ and $Q=I_n$, and consider the sequence of integer row
and column operations transforming $M$ into~$N$.  Whenever a row operation is
performed in this sequence, apply the same row operation to $P$.  Similarly,
whenever a column operation is made, apply the same column operation to $Q$.
\medskip

{\bf Exercise.}
  Explain why the resulting matrices $P$ and $Q$ are invertible over the integers and
  why $PMQ=N$.  The converse of this statement is also true: given any matrices $P$ and $Q$,
  invertible over the integers and such that $PMQ=N$, it follows that $M\sim N$.
  However, the proof of this converse requires the existence of the Smith normal
  form.
\medskip

The relation $PMQ=N$ can be expressed in terms of a commutative diagram with
exact rows:
\begin{equation}\label{cd: induced map}
  \begin{tikzcd}
    \Z^n \ar[r,"M"] \ar[d,"{Q^{-1}}"',"{\rotatebox{270}{$\approx$}}"]
    &\Z^m \ar[r] \ar[d,"{\rotatebox{90}{$\approx$}}"',"P"]
  &\cok{M} \ar[r] \ar[d,dashed]&0\\
  \Z^n \ar[r,"N"] &\Z^m \ar[r] &\cok{N} \ar[r] &0.
  \end{tikzcd}
\end{equation}
\noindent The mapping $\cok(M)\to\cok(N)$ is induced by $P$.
\medskip

{\bf Proposition.}
  Let $M$ and $N$ be $m\times n$ integer matrices.  Then if $M\sim N$, it
  follows that $\cok(M)\simeq\cok(N)$.
\smallskip

{\bf Proof.} Since $P$ and $Q$ in the commutative diagram are isomorphisms the
mapping of cokernels induced by $P$ is an isomorphism.  \hfill$\Box$
\medskip

{\bf Exercise.}
  Suppose $M\sim N$ where $N=\mathrm{diag}(m_1,\dots,m_{\ell})$, a diagonal
  integer matrix with nonnegative entries.  Show that
  \[
    \cok(M)\simeq \prod_{i=1}^{\ell}\Z/m_i\Z.
  \]
\medskip


The previous exercise shows that to determine the structure of $\cok(M)$, we should
seek to transform~$M$ through integer row and column operations into a diagonal
matrix of a particularly nice form.
\medskip

{\bf Definition.}
  An $m\times n$ integer matrix $M$ is in {\em Smith normal form} if
  \[
    M=\diag(s_1,\dots,s_k,0,\dots,0),
  \]
  a diagonal matrix, where $s_1,\dots,s_k$ are positive integers such that
  $s_i|s_{i+1}$ for all $i$.  The $s_i$ are called the {\em invariant factors}
  of $M$.
\medskip

{\bf Example.} The matrix 
\[
M:=
\left(
\begin{array}{cccc}
  1&0&0&0\\
  0&2&0&0\\
  0&0&12&0\\
  0&0&0&0\\
  0&0&0&0
\end{array}
\right)
\]
is in Smith normal form with invariant factors $s_1=1$, $s_2=2$, and $s_3=12$.  

We have 
\[
 \cok(M):= \Z^5/\im(M)\simeq \Z/1\Z\times\Z/2\Z\times\Z/{12}\Z\times\Z^2\simeq
 \Z_2\times\Z_{12}\times\Z^2.\\
\]
So $\cok(M)$ has rank $r=2$ and its invariant factors are $2$ and $12$.  

Note that $1$ is an invariant factor of $M$ but not of $\cok(M)$.  By
definition, the invariant factors of a finitely generated abelian group are
greater than $1$; the invariant factors of $M$ equal to $1$ do not affect the
isomorphism class of $\cok(M)$ since $\Z_1$ is the trivial group.
\medskip

{\bf Aside on~$\Z$-modules.} A $\Z$-module is a triple~$(M,+,\cdot)$
where~$+\colon M\times M\to M$ and~$\cdot\colon \Z\times M\to\Z$ satisfy the
usual rules for a vector space.  The difference is that the scalars are the
ring~$\Z$ rather than a field.  It is immediate to see that a~$\Z$-module is the
same thing as an abelian group since if~$A$ is an abelian group and~$n\in\Z$,
then~$na$ is well-defined as repeated addition of~$a$ with itself.

Let~$M$ be a~$\Z$-module.  Then~$M$ is {\em free} if it has a basis---a spanning
set that is linearly independent.  Not every~$\Z$-module has a basis.  For
example,~$M=\Z/5\Z$ has no basis. For instance, although~$\Z/5\Z$ is generated
by~$1$, we have~$5\cdot 1 =0$ as a non-trivial linear relation. To say that~$M$
is {\em finitely generated} means that~$M$ has a finite spanning set.  We have
that~$M$ is finitely generated and free if and only if there is a~$\Z$-linear
isomorphism
\[
  M \simeq \Z^n
\]
for some~$n$.  By the structure theorem for finite abelian groups, a submodule
of a f.~g.~free~$\Z$-module is free.  A quotient of f.g.~free~$\Z$-modules is
not necessarily free.  For instance, let~$N$ be the~$\Z$-span of the
vectors~$(2,0)$ and~$(0,3)$ in~$\Z^2$.  Then~$N$ is free with basis~$(2,0)$
and~$(0,3)$, and we have the isomorphism
\begin{align*}
  N&\simeq \Z^2\\
  (2a,3b)&\mapsto (a,b).
\end{align*}
However, we also have~$N\subset\Z^2$, and~$\Z^2/N\simeq\Z/2\Z\times\Z/3\Z$.
\end{document}
