description/proof of that for 2 subgroups with different prime number orders, if generators of subgroups are commutative, product-of-prime-numbers-ordered cyclic subgroup can be constructed
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of cyclic group by element.
- The reader admits the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.
- The reader admits the proposition that any 2 different prime-number-ordered subgroups share only 1.
- The reader admits the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal.
Target Context
- The reader will have a description and a proof of the proposition that for any group and its any 2 subgroups with any different prime number orders, if any generators of the subgroups are commutative, a the-product-of-the-prime-numbers-ordered cyclic subgroup can be constructed.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the groups }\}\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
\(q\): \(\in \{\text{ the prime numbers }\}\)
\(G_p\): \(\in \{\text{ the } p \text{ -ordered subgroups of } G\}\), \(= \{g_p, ..., g_p^p = 1\}\)
\(G_q\): \(\in \{\text{ the } q \text{ -ordered subgroups of } G\}\), \(= \{g_q, ..., g_q^q = 1\}\)
\(G_{p q}\): \(= \{g_p g_q, ..., (g_p g_q)^{p q}\}\)
//
Statements:
(
\(p \neq q\)
\(\land\)
\(g_p g_q = g_q g_p\)
)
\(\implies\)
(
\(G_{p q} \in \{\text{ the } (p q) \text{ -orderd cyclic subgroups of } G\}\)
\(\land\)
\(\forall j, k \in \mathbb{N} \text{ such that } 1 \le j \lt p \text{ and } 1 \le k \lt q (\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} = G_{p q})\)
)
//
2: Note
\(G_p\) inevitably takes the shape of \(\{g_p, ..., g_p^p = 1\}\), by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group. It is likewise for \(G_q\).
By the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, \(G_p\) or \(G_q\) can be regarded to be generated by any \(g_p^j\) or \(g_q^k\) where \(1 \le j \lt p\) and \(1 \le k \lt q\), but \(g_p^j g_q^k = g_q^k g_q^j\) holds anyway, because \(g_p^j g_q^k = g_p^{j - 1} g_p g_q g_q^{k - 1} = g_p^{j - 1} g_q g_p g_q^{k - 1} = g_p^{j - 2} g_p g_q g_p g_q^{k - 1} = g_p^{j - 2} g_q g_p g_p g_q^{k - 1} = g_p^{j - 2} g_q g_p^2 g_q^{k - 1} = ... = g_q g_p^j g_q^{k - 1} = g_q^2 g_p^j g_q^{k - 2} = ... = g_q^k g_p^j\).
The 2nd half of Statements is saying that we cannot have any different \((p q)\)-ordered cyclic subgroup by just choosing a different generator: this proposition says that \(g_p^j g_q^k\) generates a \((p q)\)-ordered cyclic subgroup, but in fact, the subgroup is no different from \(G_{p q}\).
3: Proof
Whole Strategy: Step 1: see that \(G_{p q}\) is a \((p q)\)-ordered cyclic subgroup; Step 2: see that \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} = G_{p q}\).
Step 1:
Let us see that \(G_{p q}\) is a \((p q)\)-ordered cyclic subgroup.
\(1 \in G_{p q}\): \((g_p g_q)^{p q} = g_p^{p q} g_q^{p q} = (g_p^p)^q (g_q^q)^p = 1^q 1^p = 1\).
For each \((g_p g_q)^j, (g_p g_q)^k \in G_{p q}\), \((g_p g_q)^j (g_p g_q)^k = (g_p g_q)^{j + k}\). \(j + k = m p q + l\) where \(m, l \in \mathbb{N}\) and \(0 \le l \lt p q\). \((g_p g_q)^{j + k} = (g_p g_q)^{m p q + l} = (g_p g_q)^{m p q} (g_p g_q)^l = ((g_p g_q)^{p q})^m (g_p g_q)^l = (g_p^{p q} g_q^{p q})^m (g_p g_q)^l = ((g_p^p)^q (g_q^q)^p)^m (g_p g_q)^l = (1^q 1^p)^m (g_p g_q)^l = 1^m (g_p g_q)^l = (g_p g_q)^l\). When \(1 \le l \lt p q\), \((g_p g_q)^l \in G_{p q}\) and when \(l = 0\), \((g_p g_q)^0 = 1 = (g_p g_q)^{p q} \in G_{p q}\). So, \((g_p g_q)^j (g_p g_q)^k \in G_{p q}\) anyway.
For each \((g_p g_q)^j \in G_{p q}\), \((g_p g_q)^{p q - j} \in G_{p q}\), and \((g_p g_q)^j (g_p g_q)^{p q - j} = (g_p g_q)^{p q} = 1\) and \((g_p g_q)^{p q - j} (g_p g_q)^j = (g_p g_q)^{p q} = 1\).
The associativity holds because it holds in the ambient \(G\).
So, \(G_{p q}\) is a subgroup of \(G\).
Let us see that \(\{g_p g_q, ..., (g_p g_q)^{p q}\}\) is distinct.
Let us suppose that \((g_p g_q)^j = (g_p g_q)^k\) where \(1 \le j \le k \le p q\) without loss of generality. \(1 = (g_p g_q)^{k - j}\), \(g_p^{k - j} = g_q^{j - k}\), because \(g_p g_q = g_q g_p\), \(= 1\) by the proposition that any 2 different prime-number-ordered subgroups share only 1, so, \(k - j = m p = n q\), which means that \(k - j = 0\): \(k - j\) is a multiple of \(p q\) but \(k - j \lt p q\). So, \(j = k\).
So, \(G_{p q}\) is a \((p q)\)-ordered cyclic subgroup.
Step 2:
Let us see that \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} = G_{p q}\).
\(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\}\) is a \((p q)\)-ordered cyclic subgroup by Step 1, because also \(g_p^j\) generates \(G_p\) and also \(g_q^k\) generates \(G_q\), by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, and Step 1 applies for \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\}\) with just \(g_p\) and \(g_q\) replaced with \(g_p^j\) and \(g_q^k\) respectively.
Let us see that for each \(r, s \in \mathbb{N}\) such that \(1 \le r \le p\) and \(1 \le s \le q\), \(g_p^r g_q^s \in G_{p q}\), because then, as each element of \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\}\) is of that form, it will be that \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} \subseteq G_{p q}\).
There are some \(n, m \in \mathbb{Z}\) such that \(r - s = n p + m q\) by the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal, because \(\mathbb{Z}\) is a principal integral domain, \(gcd (p, q) = \{1\}\), and \(1 \mathbb{Z} = p \mathbb{Z} + q \mathbb{Z}\).
Let \(l := r - n p = s + m q\). Then, \((g_p g_q)^l = g_p^l g_q^l = g_p^{r - n p} g_q^{s + m q} = g_p^r g_p^{- n p} g_q^s g_q^{m q} = g_p^r (g_p^p)^{- n} g_q^s (g_q^q)^m = g_p^r 1^{- n} g_q^s 1^m = g_p^r g_q^s\).
So, \(g_p^r g_q^s \in G_{p q}\).
So, \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} \subseteq G_{p q}\).
As \(\vert \{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} \vert = \vert G_{p q} \vert\) as finite sets, \(\{g_p^j g_q^k, ..., (g_p^j g_q^k)^{p q}\} = G_{p q}\).
