Application to finitely generated Abelian groups

Let us specialize to the P.I.D. $Z$. A $Z$-module is nothing but an abelian group. Unfortunately, customarily, the concepts of orders in Group Theory are in conflict with the definitions we provided for the order of any element in a module and the order of a torsion module itself. Here, we will clarify these differences. In Group Theory, the order of a group $G$, denoted by $\,\rule[-.05in]{0.005in}{.2in}\,\! G\!\,\rule[-.05in]{0.005in}{.2in}\,$, is defined to be the cardinality of the group $G$ as a set. And the order of an element $g \in G$, denoted by $\mid\!\!g\!\!\mid$, is defined to be the cardinality of the subgroup $<\!\!\{g\}\!\!>$ generated by the single element $g$. If $s = \mid\!\!g\!\!\mid < \infty$, this is equivalent to the smallest integer $s$ so that $g^{s}= g \cdots g = e$. In case of abelian group, $s \cdot g = g + \cdots + g = 0$. Hence, in the case of abelian group, if $\mid\!\!g\!\!\mid < \infty$, $\mid\!\!g\!\!\mid = o(g)$. If $\mid\!\!g\!\!\mid = \infty$, $o(g)$ is $0$ by definition. It is well-known that if $\,\rule[-.05in]{0.005in}{.2in}\,\! G\!\,\rule[-.05in]{0.005in}{.2in}\,< \infty$, for any $g \in G$, $g^{\,\rule[-.05in]{0.005in}{.2in}\,\!G\!\,\rule[-.05in]{0.005in}{.2in}\,} = e$. In case of abelian group, ${\,\rule[-.05in]{0.005in}{.2in}\,\!G\!\,\rule[-.05in]{0.005in}{.2in}\,} \cdot g = 0$. In this situation, i.e. $\,\rule[-.05in]{0.005in}{.2in}\,\! G\!\,\rule[-.05in]{0.005in}{.2in}\,< \infty$, we know for sure that $o(G) \,\rule[-.07in]{0.006in}{.25in}\, \,\rule[-.05in]{0.005in}{.2in}\,\! G\!\,\rule[-.05in]{0.005in}{.2in}\,$ and $o(G) = \,\rule[-.05in]{0.005in}{.2in}\,\! G\!\,\rule[-.05in]{0.005in}{.2in}\,$ if and only if $G$ is cyclic.

With the convention stated in the last paragraph, as a corollary of the last theorem, we have the following important structural theorem for finitely generated abelian groups:

Theorem 93   (Fundamental Theorem for Finitely Generated Abelian
Groups)

Let $G$ be a finitely generated abelian group. $G$ is a direct sum of finitely many cyclic subgroups. More precisely, there exist $m$ ( $0 \leq m \leq n$) finite cyclic subgroups $G_{1}, G_{2}, \ldots, G_{m}$ of $G$ and $n -m$ infinite cyclic subgroups $G_{m + 1}, G_{m + 2}, \ldots, G_{n}$ of $G$ such that $G = \raisebox{.03in}{${\mathrel{\mathop{\oplus}\limits^{n}\limits_{i=1}}}$} G_{i}$ and for $i = 1, 2, \ldots, m-1$, $\,\rule[-.05in]{0.005in}{.2in}\,\! G_{i + 1}\!\,\rule[-.05in]{0.005in}{.2in}\, ... ...n}\, \,\rule[-.05in]{0.005in}{.2in}\,\! G_{i}\!\,\rule[-.05in]{0.005in}{.2in}\,$.

Corollary:

Any finitely generated abelian group $A$ is torsion if and only if $\,\rule[-.05in]{0.005in}{.2in}\,\!A\!\,\rule[-.05in]{0.005in}{.2in}\,< \infty$.

Examples:

1. Analyze the abelian group $G$ with $\,\rule[-.05in]{0.005in}{.2in}\,\!G\!\,\rule[-.05in]{0.005in}{.2in}\,= 64$.

   Solution:

   The well-known fact: If $G = H \oplus K$, $\,\rule[-.05in]{0.005in}{.2in}\,\!G\!\,\rule[-.05in]{0.005in}{.2in}\,= \,\rule[... ...\, \cdot \,\rule[-.05in]{0.005in}{.2in}\,\!K\!\,\rule[-.05in]{0.005in}{.2in}\,$ will be helpful. We have the following few cases to consider:

   $o(G) = 2$: $G = Z_{2} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2}$.

   $o(G) = 4$: $G = Z_{4} \oplus Z_{4} \oplus Z_{4}$, $G = Z_{4} \oplus Z_{4} \oplus Z_{2} \oplus Z_{2}$, or

   $G = Z_{4} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2}$.

   $o(G) = 8$: $G = Z_{8} \oplus Z_{8}$, $G = Z_{8} \oplus Z_{4} \oplus Z_{2}$, or $G = Z_{8} \oplus Z_{2} \oplus Z_{2} \oplus Z_{2}$

   $o(G) = 16$: $G = Z_{16} \oplus Z_{4}$ or $G = Z_{16} \oplus Z_{2} \oplus Z_{2}$.

   $o(G) = 32$: $G = Z_{32} \oplus Z_{2}$

   $o(G) = 64$: $G = Z_{64}$

   Exercise: Analyze the abelian group $G$ with $\,\rule[-.05in]{0.005in}{.2in}\,\!G\!\,\rule[-.05in]{0.005in}{.2in}\,= 96$.

2. Let us reconsider the example on Page .

   $\left[\!\!\begin{array}{rrr} 1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6\\ 0 & 0 &... ...{array}{rrr} 1 & -1 & 0\\ 0 & 1 & 0\\ 1 & -1 & 1 \end{array} \!\!\right]$

   In ${\cal Z}$-modules $Z^{3}$ and $Z^{4}$, we choose $\{<\!\! 1, 0, 1\!\!>, <\!\! -1, 1, -1\!\!>, <\!\! 0, 0, 1\!\!>\}$ and $\{<\!\! 1, 2, 1, 1\!\!>, <\!\! 3, 5, 3, 3\!\!>, <\!\! 1, 2, 8, 2\!\!>, <\!\! 1, 2, 9, 2\!\!>\}$ as their bases, respectively. Shorten the notations by $\{v_{1}, v_{2}, v_{3}\}$ and $\{w_{1},w_{2},w_{3},w_{4}\}$. It is easy to verify: $\alpha_{_{M}}(v_{1}) = 1 \cdot w_{1}$, $\alpha_{_{M}}(v_{2}) = 3 \cdot w_{2}$, $\alpha_{_{M}}(v_{3}) = 6 \cdot w_{3}$. Hopefully, this will clarify that $\{1, 3, 6\}$ is the scalars for $\alpha_{_{M}}(Z^{3}) \leq Z^{4}$.
   

   Exercise: Find the T-decomposition for $Z^{4}/\alpha_{_{M}}(Z^{3})$.

3. Given an $n$-dimensional vector space $V$ and an operator $\delta \in {\cal L}(V, V)$, in Section 7.2, we have learned how to turn $V$ and $\delta$ into a finitely generated torsion $K[x]$-module: $<\!\!V_{\delta}, +, K[x], \bullet_{\delta}\!\!>$. Conversely, given any finitely generated torsion $K[x]$-module $M_{t} = <\!\!M_{t}, +, K[x], \cdot\!\!>$, we can induce from it a finite dimensional space $V_{M_{t}}$ and a linear operator $\alpha_{x} \in {\cal L}(V_{M_{t}}, V_{M_{t}})$ as follows:

   The vector space $<\!\!V_{M_{t}}, +, K, \cdot\!\!>$:

   As a set, we define $V_{M_{t}} = M_{t}$. The addition operation "+" and scalar multiplication "$\cdot$" are inherited from the $K[x]$-module $M_{t}$:

   $\begin{array}{cccc} \begin{array}{ccccc} + & : & V_{M_{t}} \times V_{M_{t}} &... ... & w & \mapsto & x \cdot w \\ & & & & \\ & & & & \end{array} \end{array}$

   Exercise: Verify that $<\!\!V_{M_{t}}, +, K, \cdot\!\!>$ is indeed a vector space.

   The operator $\alpha_{x} \in {\cal L}(V_{M_{t}}, V_{M_{t}})$: $\alpha_{x}$ is linear, since

   $\alpha_{x}(v + w) = x \cdot (v + w) = x \cdot v + x \cdot w = \alpha_{x}(v) + \alpha_{x}(w)$ and $\alpha_{x}(c \cdot w) = x \cdot c \cdot w = c \cdot x \cdot w = c \alpha_{x}(w)$.

   Since $M_{t}$ is a finitely generated torsion $K[x]$-module, by Theorem 92, $M_{t} = K[x]{w_{1}} \bigoplus \cdots \bigoplus K[x]{w_{k}}$ with reversed order. Let us assume that
   

   
   

   $$\begin{array}{ccl} o(w_{1}) & = & x^{n_{1}} + a_{(n_{1}-1,1)... ...n_{k} -1} + \cdots + a_{(1,k)}\cdot x + a_{(0,k)} \end{array}$$
   
   

   Since $o(w_{i}) \,\,\rule[-.05in]{0.005in}{.2in}\,\, o(w_{i+1})$, $1 \leq i \leq k -1$, we know that $n_{1} \leq n_{2} \leq \cdots \leq n_{k}$. By observing that $M_{t} = K[x]{w_{1}} \bigoplus \cdots \bigoplus K[x]{w_{k}}$, we see that vectors in $K[x]{w_{i}}$ should be independent of vectors in $K[x]{w_{j}}$, for $i \not= j$. And $o(w_{i})$ is the minimal polynomial for combining those vectors in the set $\{w_{i}, x \cdot w_{i},\ \ldots\ , x^{n_{i}} \cdot w_{i}\}$ to get $0$. It is easy to show that the following set is a basis for $V_{M_{t}}$ and its dimension is hence $n_{1} + n_{2} + \cdots + n_{k}$:

   $\{w_{1}, x \cdot w_{1},\ \ldots\ , x^{n_{1} -1} \cdot w_{1}, w_{2}, x \cdot w_{2},\ \ldots\ , w_{k}, x \cdot w_{k},\ \ldots\ , x^{n_{k} -1} \cdot w_{k}\}$

   Also, observe that $o(w_{k}) = x^{n_{k}} + a_{(n_{k}-1,k)} \cdot x^{n_{k} -1} + \cdots + a_{(1,k)}\cdot x + a_{(0,k)}$ is the minimal polynomial of the linear operator $\alpha_{x}$, since it annihilates each vector in $\{w_{1}, w_{2}, \ldots, w_{k}\}$. (The fact that $o(w_{i}) \,\,\rule[-.05in]{0.005in}{.2in}\,\, o(w_{k})$ for $i \leq k$ is useful, here.) What is not so trivial is that $o(w_{1}) \cdot o(w_{2}) \cdots o(w_{k})$ is the characteristic polynomial for $\alpha_{x}$. Applying Definition 66.2, it won't be too hard to construct a rational canonical matrix to represent $\alpha_{x}$.

Exercises:

1. Look at the following computation:

   
   

   $$\hspace*{-.2in}\begin{array}{cccccccc} \left[\!\!\begin{arra... ...& 1 \\ \ 0 & 0 & 1 & 0 \end{array} \!\!\right] \end{array}$$
   
   
   1. Find a basis for the submodule $<\!\!\{(1, 0, -1), (2, 0, 2), (0, 3, 1), (3, 1, 5)\}\!\!>$ of the free module $Z \oplus Z \oplus Z$.

   2. Let $\alpha : Z^{3} \longrightarrow Z^{4}$ be the $Z$-homomorphism induced by $\left[\!\!\begin{array}{rrrr} 1 & 0 & \!-1 \\ 2 & 0 & \! 2 \\ 0 & 3 & \! 1 \\ 3 & 1 & \! 5 \end{array} \!\!\right]$, identify $Z^{4}/\alpha(Z^{3})$. Find the invariant factors for $Z^{4}/\alpha(Z^{3})$.

      Suggestion: Probably, it will be easier after you finish the next section. At least, you can use the contents of the next section to confirm your answer.

2. Let $0 \longrightarrow Y \stackrel{\mu}{\longrightarrow} X \stackrel{\nu} {\longrightarrow} Z \longrightarrow 0$ be any short exact sequence of finitely generated ${\cal D}$-modules, where ${\cal D}$ is a principle ideal domain, show that ${\rm rank}(X) = {\rm rank}(Y) + {\rm rank}(Z)$
   

   Remark:

   In Algebraic Topology, ${\rm rank}(X)$ is called the Betti number of $X$, and it is an important topological invariant which measures the components of a topological space.

3. For the lemma on Page  and the proof of part 2 of Theorem 92, this exercise provides them a useful example:
   1. Identify ${\rm Ker}_{{\cal Z}_{72} \oplus {\cal Z}_{25}}$ of the following short exact sequence:

      $0 \longrightarrow {\rm Ker}_{{\cal Z}_{72} \oplus {\cal Z}_{25}} \longrightarr... ...ghtarrow} 3 \cdot {\cal Z}_{72} \oplus 3 \cdot {\cal Z}_{25} \longrightarrow 0$

   2. Let ${\cal D}_{w}$ be a cyclic module generated by $w$, $a = o(w)$, and $p$ be a prime divisor of $a$. Show that ${\rm Ker}_{{\cal D}_{w}} = {\cal D} / (p)$ where ${\rm Ker}_{{\cal D}_{w}}$ is the second term of: $0 \longrightarrow {\rm Ker}_{{\cal D}_{w}} \longrightarrow {\cal D}_{w} \stackrel{\textstyle{\times p}}{\longrightarrow} p \cdot {\cal D}_{w} \longrightarrow 0$.
4. Find all non-isomorphic Abelian groups with 72 elements.