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Let \(G_1,\cdots,G_n\) be groups. For \((a_1,\cdots,a_n)\) and \((b_1,\cdots,b_n)\) in \(\prod_{i=1}^nG_i\), define \((a_1,\cdots,a_n)(b_1,\cdots,b_n) = (a_1b_1,\cdots,a_nb_n)\).
定義 element-wise operation。
The group \(\prod_{i=1}^nG_i = G_1 \times \cdots \times G_n\) is called the direct product of the groups \(G_i\).
The group \(\mathbb{Z}_m \times \mathbb{Z}_n\) is cyclic and isomorphic to \(\mathbb{Z}_{mn}\) if and only if \(m\) and \(n\) are relatively prime.
Proof
(\(\Rightarrow\))
Suppose \(\mathbb{Z}_m \times \mathbb{Z}_n\) is cyclic and has a generator \((\bar a, \bar b)\), which is of order \(mn\). Let \(k = \text{lcm}(m,n)\). Then
\[k(\bar a, \bar b) = (\overline{ka}, \overline{kb}) = (\bar 0, \bar 0).\]\(Z_m\) 的 generator is of order \(m\)。
Therefore \(mn \le k = \text{lcm}(m,n)\), which means that \(m\) and \(n\) are relatively prime.
(\(\Leftarrow\))
Conversly, assume that \(m\) and \(n\) are relatively prime, which means \(\text{lcm}(m,n)=mn\). Let the order of the element \((\bar 1, \bar 1)\) to be \(k\). Then \(k \le mn\) and
\[(\bar 0,\bar 0) = k(\bar 1, \bar 1) = (\bar k, \bar k).\]Therefore, \(m \vert k\) and \(n \vert k\), which means \(\text{lcm}(m,n) \vert k\). In particular, we have
\[mn = \text{lcm}(m,n) \le k \le mn.\]We conclude that \(k = mn\), \((\bar 1, \bar 1)\) is a generator, and \(\mathbb{Z}_m \times \mathbb{Z}_n\) is cyclic. ◼
Let \((a_1,\cdots,a_n) \in \prod^n_{i=1}G_i\). If \(a_i\) is of finite order \(r_i\) in \(G_i\), then the order of \((a_1,\cdots,a_n)\) is equal to \(\text{lcm}(r_1,\cdots,r_n)\).
用特定 element(s) 的 order 或 group 是否 cyclic 來判斷兩 groups 是否 isomorphic！
當然，如果 groups 的 order 不同，也不可能存在 isomorphism！
e.g.
max element order:
\[\begin{align*} \mathbb{Z}_8 &: 8 \\ \mathbb{Z}_{4}\times\mathbb{Z}_2 &: 4 \\ \mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2 &: 2 \end{align*}\]所以以上三 groups 兩兩不 isomorphic。
An abelian group is finitely generated if it can be generated by a finite number of elements.
Suppose that \(G\) is a finitely generated abelian group. Then \(G\) is isomorphic to a direct product of cyclic groups in the form
\[\mathbb{Z}_{p_1^{e_1}}\times\mathbb{Z}_{p_2^{e_2}}\times\cdots\mathbb{Z}_{p_n^{e_n}}\times\mathbb{Z}\times\cdots\times\mathbb{Z},\]where \(p_i\) are primes, not necessarily distinct. The direct product is unique except for possible rearrangement of the factors.
後方 \(\mathbb{Z}\) 的數量稱為 the free rank of \(G\)。
e.g.
Using the above theorem, we find that the only abelian groups of order \(360=2^33^25\), up to isomorphism, are
\[\begin{align*} &\mathbb{Z}_{2^3}\times\mathbb{Z}_{3^2}\times\mathbb{Z}_{5}\ (\cong \mathbb{Z}_{360}), \\ &\mathbb{Z}_{2^2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3^2}\times\mathbb{Z}_{5}, \\ &\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3^2}\times\mathbb{Z}_{5}, \\ &\mathbb{Z}_{2^3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{5}, \\ &\mathbb{Z}_{2^2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{5}, \\ &\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{5}. \end{align*}\]In general, if the prime factorization of a positive integer \(n\) is \(n=p_1^{e_1}\cdots p_k^{e_k}\), where \(p_i\) are distinct primes, then the number of all abelian groups of order \(n\), up to isomorphism, is equal to
\[p(e_1)\cdots p(e_k),\]where \(p(e)\) is the number of ways to partition the integer \(e\).
上述例子共 \(3\cdot 2 \cdot 1 = 6\) 個 abelian groups。
If \(m\) divides the order of a finite abelian group \(G\), then \(G\) has a subgroup of order \(m\).
Proof
By the above theorem, we have \(G=\mathbb{Z}_{p_1^{e_1}}\times\mathbb{Z}_{p_2^{e_2}}\times\cdots\mathbb{Z}_{p_n^{e_n}}\).
If \(m \Big\vert \vert G \vert\), then \(m = p_1^{f_1}\cdots p_n^{f_n}\) for some non-negative integers \(f_i\). (This representation may not be distinct.)
Then the subgroup
\[\langle p_1^{e_1-f_1} \rangle \times \cdots \times \langle p_n^{e_n-f_n} \rangle\]has order \(p_1^{f_1}\cdots p_n^{f_n} = m\). ◼
\(\overline{p_1^{e_1-f_1}} \cdot p_1^{f_1} = \overline{p_1^{e_1}} = \bar 0\), under \(\mathbb{Z}_{p_1^{e_1}}\).
連加 \(p_1^{f_1}\) 次就回到 identity！
Follow from the fundamental theorem, we can choose any \(p = p_i\) to obtain a cyclic subgroup \(G_p\), characterized by
\[G_p = \{a\in G \vert a^{p^k} = e \text{ for some } k \in \mathbb{N} \},\]and \(G_p\) is called the \(p\)-torsion subgroup of \(G\).
We can rephrase the above result as
\[G = \prod_{p\big\vert \vert G \vert} G_p,\]where \(G_p\) is isomorphic to a group of the form
\[\mathbb{Z}_{p^{r_1}}\times \cdots \mathbb{Z}_{p^{r_k}}.\]Therefore, to determine the group structure of \(G\), it is sufficient to determine the group structure of its p-torsion subgroups.
For a positive integer \(n\), consider the following subsets
\[G^{(n)} = \{a^n\vert a\in G \},\]which is a subgroup.
Example
For a group \(G\) of order \(16\), it can be isomorphic to one of the four groups:
\[\mathbb{Z}_{16}, \mathbb{Z}_{8}\times \mathbb{Z}_{2}, \mathbb{Z}_{4}\times \mathbb{Z}_{4}, \mathbb{Z}_{4}\times (\mathbb{Z}_{2})^2, (\mathbb{Z}_{2})^4\]After computing these subsets of each posssible structure of \(G\), we obtain
| \(G\) | \(\mathbb{Z}_{16}\) | \(\mathbb{Z}_{8}\times \mathbb{Z}_{2}\) | \(\mathbb{Z}_{4}\times \mathbb{Z}_{4}\) | \(\mathbb{Z}_{4}\times (\mathbb{Z}_{2})^2\) | \((\mathbb{Z}_{2})^4\) |
|---|---|---|---|---|---|
| \(G^{(2)}\) | \(\mathbb{Z}_{8}\) | \(\mathbb{Z}_{4}\) | \(\mathbb{Z}_{2}\times \mathbb{Z}_{2}\) | \(\mathbb{Z}_{2}\) | \(\{e\}\) |
| \(G^{(4)}\) | \(\mathbb{Z}_{4}\) | \(\mathbb{Z}_{2}\) | \(\{e\}\) | \(\{e\}\) | \(\{e\}\) |
which implies the following results:
以此類推，用以上表格就可以唯一決定 \(G\) 的 structure！
Note: For a group \(T \cong \mathbb{Z}_{mn}\), we have \(T^{(n)} \cong n\mathbb{Z}_{mn}\cong \mathbb{Z}_m\). 也就是說，\(n\mathbb{Z}_n = \{e\}\)，消失了！
綜合以上，對於任意 group \(G\)，我們可以用一演算法來決定其結構：
關於 cardinality，補一個 theorem：
\(\vert \mathbb{Z}_m^{\times} \vert = m\prod_p(1-1/p)\), where \(p\) runs through all prime divisors of \(m\).