Finitely Generated Abelian Groups Structure theorem的大致内容就是任一有限 Abel 群都同构于若干个循环群的直和.

（1） $G \simeq \bigoplus_{i=1}^t\left(\bigoplus_{j=1}^{k_i} Z_{p_i}\right)$ ，其中 $l_{i j}$ 是一组正整数，满足 $\sum_{j=1}^{k_i} l_{i j}=e_i$ ；
(2) $G \simeq \bigoplus_{i=1}^k Z_{d_i}$ ，其中 $d_i$ 是一组正整数，满足 $d_1\left|d_2\right| \cdots \mid d_k$ 且 $\prod_{i=1}^k d_i=n$. 这些 $d_i$ 称为 $G$ 的不变因子.

UpriviateTA有若干位非常擅长Abstract algebra的老师可以提供抽象代数的代写，代考和辅导服务，可以确保您在学习Abstract algebra的过程中取得满意的成绩。

Problem 1.

Let $\alpha$ be a partition of an integer and denote by $G_\alpha$ the group
$$G_\alpha=Z_{2^{\alpha_1}} \oplus Z_{2^{\alpha_2}} \oplus \cdots \oplus Z_{2^{\alpha_{\ell(\alpha)}}} .$$
(a) Prove that $G_\alpha$ contains $2^{\ell(\alpha)}-1=2^{\alpha_1^{\prime}}-1$ elements of order 2 .
(b) Prove that $G_\alpha$ contains $2^{|\alpha|}-2^{|\alpha|-m}$ elements of order $2^{\alpha_1}$, where $\alpha_1$ appears $m$ times in the partition $\alpha$.

Problem 2.

The Exponent of Matrix Groups. Fermat’s Little Theorem can be rephrased as: if $p$ is a prime, then $\forall \bar{a} \in U(p), \bar{a}^{p-1}=\overline{1}$. This is an immediate consequence of Lagrange’s Theorem. The exponent of a group $G$, denoted $\exp (G)$, is the least positive integer $m$ such that $g^m=1$ for $g \in G$. Explore the exponent $k(n, p)$ of the group $G=\mathrm{GL}_n\left(\mathbb{F}_p\right)$.

## Free abelian groups

The Fundamental Theorem of Finitely Generated Abelian Groups (abbreviated by FTFGAG), which, among other things, provides a complete classification of all finite abelian groups. The proof begins with a study of free abelian groups.

A subset $X \subseteq G$ of an abelian group is called linearly independent if for every finite subset $\left{x_1, x_2, \ldots, x_r\right} \subseteq X$,
$$c_1 x_1+c_2 x_2+\cdots+c_r x_r=0 \Longrightarrow c_1=c_2=\cdots=c_r=0 .$$
A basis of an abelian group $G$ is a linearly independent subset $X$ that generates $G$.

free abelian group

An abelian group $(G,+)$ is called a free abelian group if it has a basis.

In particular, $\mathbb{Z}, \mathbb{Z} \oplus \mathbb{Z}$, and more generally $\mathbb{Z}^r$ for a positive integer $r$ are free abelian groups. A free abelian group could have an infinite basis. For example, $\mathbb{Z}[x]$ is a free abelian group with basis $S=\left{1, x, x^2, \ldots\right}$ because every polynomial in $\mathbb{Z}[x]$ is a (finite) linear combination of elements in $S$. If a free abelian group has an infinite basis, every element must still be a finite linear combination of basis elements.

Theorem 1.

Let $G$ be a nonzero free abelian group with a basis of $r$ elements. Then $G$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}=\mathbb{Z}^r$.

Proof. Let $X=\left{x_1, x_2, \ldots, x_r\right}$ be a finite basis of $G$. Consider the function $\varphi: \mathbb{Z}^r \rightarrow G$ given by
$$\varphi\left(c_1, c_2, \ldots, c_r\right)=c_1 x_1+c_2 x_2+\cdots+c_r x_r .$$
This function satisfies
\begin{aligned} \varphi\left(\left(c_1, c_2, \ldots, c_r\right)+\left(d_1, d_2, \ldots, d_r\right)\right) \ \quad=\varphi\left(c_1+d_1, c_2+d_2, \ldots, c_r+d_r\right) \ \quad=\left(c_1+d_1\right) x_1+\left(c_2+d_2\right) x_2+\cdots+\left(c_r+d_r\right) x_r \ \quad=c_1 x_1+c_2 x_2+\cdots+c_r x_r+d_1 x_1+d_2 x_2+\cdots+d_r x_r \ \quad=\varphi\left(c_1, c_2, \ldots, c_r\right)+\varphi\left(d_1, d_2, \ldots, d_r\right) \end{aligned}
so it is a homomorphism. Since the basis generates $G$, then $\varphi$ is surjective. Furthermore, since
\begin{aligned} \operatorname{Ker} \varphi & =\left{\left(c_1, c_2, \ldots, c_r\right) \in \mathbb{Z}^r \mid c_1 x_1+c_2 x_2+\cdots+c_r x_r=0\right} \ & ={(0,0, \ldots, 0)}, \end{aligned}
the homomorphism is also injective. Thus, $\varphi$ is an isomorphism.

Theorem 2.

Let $G$ be a finitely generated free abelian group. Then every basis of $G$ has the same number of elements.

Proof. Suppose that $G$ has a basis $X$ with $r$ elements. Then $G$ is isomorphic to $\mathbb{Z}^r$. The subgroup $2 G={g+g \mid g \in G}$ is isomorphic to $(2 \mathbb{Z})^r$ so by Exercise 2.4.7,
$$G / 2 G=(\mathbb{Z} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}) /(2 \mathbb{Z} \oplus 2 \mathbb{Z} \oplus \cdots \oplus 2 \mathbb{Z}) \cong Z_2^r .$$
Thus, $|G / 2 G|=2^r$.
First, assume that $G$ also has a finite basis with $s \neq r$ elements. Then $|G / 2 G|=2^s \neq 2^r$, a contradiction.

Second, assume that $G$ has an infinite basis $Y$. Let $y_1, y_2 \in Y$. Assume also that $\bar{y}_1=\bar{y}_2$ in $G / 2 G$, then $y_1-y_2 \in 2 G$, so $y_1-y_2$ is a finite linear combination of elements in $Y$ (with even coefficients). In particular, $Y$ is a linearly dependent set, which contradicts $Y$ being a basis. Thus, in the quotient group $G / 2 G$, the elements ${\bar{y} \mid y \in Y}$ are all distinct. Since $Y$ is an infinite set, so is $G / 2 G$. This contradicts $|G / 2 G|=2^r$.

Consequently, if $G$ has a basis of $r$ elements, then every other basis is finite and has $r$ elements.

If $G$ is a finitely generated free abelian group, then the common number $r$ of elements in a basis is called the rank. The rank is also called the Betti number of $G$ and is denoted by $\beta(G)$.

Theorem 3. Let $G$ be a nonzero free abelian group of finite rank $s$ and let $H \leq G$ be a nontrivial subgroup. Then $H$ is a free abelian group of rank $t \leq s$. There exists a basis $\left{x_1, x_2, \ldots, x_s\right}$ for $G$ and positive integers $n_1, n_2, \ldots, n_t$, where $n_i$ divides $n_{i+1}$ for all $1 \leq i \leq t-1$ such that $\left{n_1 x_1, n_2 x_2, \ldots, n_t x_t\right}$ is a basis of $H$.

Proof. We prove the theorem by starting from a basis of $G$ and repeatedly adjusting it

By the well-ordering of the integers, there exists a minimum value $n_1$ in the set
$\left{c_1 \in \mathbb{N}^* \mid c_1 y_1+c_2 y_2+\cdots+c_s y_s \in H\right.$ for any basis $\left{y_1, y_2, \ldots, y_s\right}$ of $\left.G\right}$.
Let $z_1 \in H$ be an element that instantiates this minimum value, and write $z_1=n_1 y_1+c_2 y_2+\cdots+c_s y_s$. By integer division, for all $i \geq 2$, we can write $c_i=n_1 q_i+r_i$ with $0 \leq r_i<n_1$. Set $x_1=y_1+q_2 y_2+\cdots+q_s y_s$. $\left{x_1, y_2, \ldots, y_s\right}$ is a basis of $G$. Furthermore,
$$z_1=n_1 x_1+r_2 y_2+\cdots+r_s y_s .$$
However, since $n_1$ is the least positive coefficient that occurs in any linear combination over any basis of $G$ and, since $0 \leq r_i<n_1$, we have $r_2=\cdots=$ $r_s=0$. So, in fact $z_1=n_1 x_1$.

If $\left{n_1 x_1\right}$ generates $H$, then we are done and $\left{n_1 x_1\right}$ is a basis of $H$. If not, then there is a least positive value of $c_2$ for linear combinations
$$z_2=a_1 x_1+c_2 y_2+\cdots+c_s y_s \in H,$$

where $y_2, \ldots, y_s \in G$ such that $\left{x_1, y_2, \ldots, y_s\right}$ is a basis of $G$. Call this least positive integer $n_2$. Note that we must have $n_1 \mid a_1$, because otherwise, since $n_1 x_1 \in H$ by subtracting a suitable multiple of $n_1 x_1$ from $a_1 x_1+c_2 y_2+\cdots+$ $c_s y_s$ we would obtain a linear combination of $\left{x_1, y_2, \ldots, y_s\right}$ that is in $H$ and has a lesser positive coefficient for $x_1$, which would contradict the minimality of $n_1$. Again, if we take the integer division of $c_i$ by $n_2, c_i=n_2 q_i+r_i$ with $0 \leq r_i0$ would contradict the minimality of $n_2$. Thus, $z_2=a_1 x_1+n_2 x_2 \in H$ and also $n_2 x_2=z_2-\left(a_1 / n_1\right) n_1 x_1 \in H$. By Lemma 2.5.8, $\left{x_1, x_2, y_3, \ldots, y_s\right}$ is a basis of $G$. Furthermore, if we consider the integer division of $n_2$ by $n_1$, written as $n_2=n_1 q+r$ with $0 \leq r<n_1$, then
$$n_1 x_1+n_2 x_2=n_1\left(x_1+q x_2\right)+r x_2 \in H .$$
But then, since $r<n_1$, by the minimal positive condition on $n_1$, we must have $r=0$. Hence, $n_1 \mid n_2$.

If $\left{n_1 x_2, n_2 x_2\right}$ generates $H$, then we are done because the set is linearly independent since by construction $\left{x_1, x_2, y_3, \ldots, y_s\right}$ is a basis of $G$ so $\left{x_1, x_2\right}$ is linearly independent and $\left{n_1 x_1, n_2 x_2\right}$ is then a basis of $H$. The pattern continues and only terminates when it results in a basis
$$\left{x_1, \ldots, x_t, y_{t+1}, \ldots, y_s\right}$$
of $G$ such that $\left{n_1 x_1, n_2 x_2, \ldots, n_t x_t\right}$ is a basis of $H$ for some positive integers $n_i$ such that $n_i \mid n_{i+1}$ for $1 \leq i \leq t-1$.

This proof is not constructive since it does not provide a procedure to find the $n_i$, which is necessary to construct the $x_1, x_2$, and so on. We merely know the $n_i$ exist by the well-ordering of integers. In some instances it is easy to find a basis of the subgroup as in the following example.

#### Example

Consider the free abelian group $G=\mathbb{Z}^3$ and the subgroup $H=\left{(x, y, z) \in \mathbb{Z}^3 \mid x+2 y+3 z=0\right}$. If we considered the equation $x+$ $2 y+3 z=0$ in $\mathbb{Q}^3$, then the Gauss-Jordan elimination algorithm gives $H$ as the span of $\operatorname{Span}({(-2,1,0),(-3,0,1)}$. Taking only integer multiples of these two vectors does give all points $(x, y, z)$ with $y$ and $z$ taking on every pair of integers. Hence, ${(-2,1,0),(-3,0,1)}$ is a basis of $H$ and we see clearly that $H$ is a free abelian group of rank 2.

## Invariant Factors Decomposition

The difficult work in the proof of above Theorem leads to the Fundamental Theorem for Finitely Generated Abelian Groups.

Theorem 4. (FTFGAG)
Let $G$ be a finitely generated abelian group. Then $G$ can be written uniquely as
$$G \cong \mathbb{Z}^r \oplus Z_{d_1} \oplus Z_{d_2} \oplus \cdots \oplus Z_{d_k}$$
for some nonnegative integers $r, d_1, d_2, \ldots, d_k$ satisfying $d_i \geq 2$ for all $i$ and $d_{i+1} \mid d_i$ for $1 \leq i \leq k-1$.

Proof. Since $G$ is finitely generated, there is a finite subset $\left{g_1, g_2, \ldots, g_s\right}$ that generates $G$. Define the function $h: \mathbb{Z}^s \rightarrow G$ by
$$h\left(n_1, n_2, \ldots, n_s\right)=n_1 g_1+n_2 g_2+\cdots+n_s g_s .$$
By the same reasoning as the proof of Theorem 2.5.5, $h$ is a surjective homomorphism. Then Ker $h$ is a subgroup of $\mathbb{Z}^s$ and by the First Isomorphism Theorem, since $h$ is surjective, $\mathbb{Z}^s /(\operatorname{Ker} h) \cong G$. By Theorem 2.5 .9 , there exists a basis $\left{x_1, x_2, \ldots, x_s\right}$ for $\mathbb{Z}^s$ and positive integers $n_1, n_2, \ldots, n_t$, where $n_i$ divides $n_{i+1}$ for all $1 \leq i \leq t-1$ such that $\left{n_1 x_1, n_2 x_2, \ldots, n_t x_t\right}$ is a basis of Ker $h$. Then
\begin{aligned} G & \cong(\mathbb{Z} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}) /\left(n_1 \mathbb{Z} \oplus \cdots \oplus n_t \mathbb{Z} \oplus{0} \oplus \cdots \oplus{0}\right) \ & \cong Z_{n_1} \oplus Z_{n_2} \oplus \cdots \oplus Z_{n_t} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z} . \end{aligned}
Now if $n_i=1$, then $Z_{n_i} \cong{0}$, the trivial group. Let $k$ be the number of indices such that $n_i>1$. The theorem follows after setting $d_i=n_{t+1-i}$ for $1 \leq i \leq k$.
Definition 2.5.12
As with free groups, the integer $r$ is called the rank or the Betti number of $G$. It is sometimes denoted by $\beta(G)$. The integers $d_1, d_2, \ldots, d_k$ are called the invariant factors of $G$ and the expression (2.7) is called the invariant factors decomposition of $G$.
It is interesting to note that the proof of Theorem 2.5.11 is not constructive in the sense that it does not provide a method to find specific elements in $G$ whose orders are the invariant factors of $G$. The invariant factors exist by virtue of the well-ordering principle of the integers.

Applied to finite groups (which obviously are finitely generated), Theorem 2.5.11 gives us an effective way to describe all abelian groups of a given order $n$. If $G$ is finite, the $\operatorname{rank}$ of $G$ is 0 . Then we must find all finite sequences of integers $d_1, d_2, \ldots, d_k$ such that

• $d_i \geq 2$ for $1 \leq i \leq k$;
• $d_{i+1} \mid d_i$ for $1 \leq i \leq k-1$;
• $n=d_1 d_2 \cdots d_k$.

The first two conditions are explicit in the above theorem. The last condition follows from the fact that $d_i=\left|x_i\right|$ where the $\left{x_1, x_2, \ldots, x_k\right}$ is a list of corresponding generators of $G$. Then every element in $G$ can be written uniquely as
$$g=\alpha_1 x_1+\alpha_2 x_2+\cdots+\alpha_k x_k$$
for $0 \leq \alpha_i<d_i$. Thus, the order of the group is $n=d_1 d_2 \cdots d_k$.
We list a few examples of abelian groups of a given order where we find the invariant factors decomposition. (Note that in the notation $Z_d$ for a cyclic group, we assume the operation is multiplication but that is irrelevant to the theorem.)

## A Few Comments on Partitions of Integers

Partitions of integers form a vast area of research and find applications in many areas of higher mathematics. See [19, Chapter XIX] or [1] for an introduction to partitions of integers.

Considering a partition $\alpha$ as any finite non-increasing sequence of positive integers, we use the notation
$$|\alpha| \stackrel{\text { def }}{=} \alpha_1+\alpha_2+\cdots+\alpha_k$$
and call this the content of $\alpha$. The integer $k$ is the length (or the number of parts) of $\alpha$ and is written $\ell(\alpha)=k$. In the terminology of partitions, each summand $\alpha_i$ of a partition is called a part of $\alpha$. For example, if $\alpha=(5,2,2,1)$, then the content of $\alpha$ is $|\alpha|=10$, so $\alpha$ is a partition of 10 with 4 parts.

We often represent a partition by its so-called Young diagram in which each part $\alpha_i$ is represented by $\alpha_i$ boxes that are left aligned, descending on the page as $i$ increases. The Young diagram for $(5,2,2,1)$ is:

Like a matrix, we call the main diagonal of the Young diagram the set of $i$ th boxes in the $i$ th row.

The conjugate of a partition $\alpha$ is the partition $\alpha^{\prime}$ obtained by reflecting the Young diagram of $\alpha$ through its main diagonal. Algebraically, the values of the conjugate partition are
$$\alpha_j^{\prime}=\left|\left{1 \leq i \leq k \mid \alpha_i \geq j\right}\right| .$$
With $\alpha=(5,2,2,1)$, the conjugate partition is $\alpha^{\prime}=(4,3,1,1,1)$. The intuition of the Young diagram makes it clear that $\left|\alpha^{\prime}\right|=|\alpha|$, so if $\alpha$ is a partition of $n$, then so is $\alpha^{\prime}$.

abstract algebra代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。