description/proof of that for group and 2 commutative elements of different orders, order of multiplication of 2 elements is least common multiple of orders
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of order of element of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any 2 commutative elements of any different orders, the order of the multiplication of the 2 elements is the least common multiple of the orders.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g_1\): \(\in G\), with \(\vert (g_1) \vert = n_1\), where \(n_1 \in \mathbb{N} \setminus \{0\}\)
\(g_2\): \(\in G\), with \(\vert (g_2) \vert = n_2\), where \(n_2 \in \mathbb{N} \setminus \{0\}\)
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Statements:
(
\(g_1 g_2 = g_2 g_1\)
\(\land\)
\(n_1 \lt n_2\)
)
\(\implies\)
\(\vert (g_1 g_2) \vert = lcm (n_1, n_2)\)
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2: Proof
Whole Strategy: Step 1: let \(l := lcm (n_1, n_2) = n_1 m_1 = n_2 m_2\) where \(gcd (m_1, m_2) = 1\); Step 2: see that \((g_1 g_2)^l = 1\) and see that \(n := \vert (g_1 g_2) \vert\) is a divisor of \(l\); Step 3: suppose that \(n \lt l\), see that \(n\) was a common divisor of \(n_1\) and \(n_2\), let \(n_j = n n'_j\), and find a contradiction from \(((g_1 g_2)^n)^{n'_1} = 1\).
Step 1:
Let \(l := lcm (n_1, n_2) = n_1 m_1 = n_2 m_2\).
\(gcd (m_1, m_2) = 1\), because otherwise, \(n_1 m_1 / gcd (m_1, m_2) = n_2 m_2 / gcd (m_1, m_2)\) would be a smaller common multiple.
Step 2:
\((g_1 g_2)^l = g_1^l g_2^l\), because \(g_1 g_2 = g_2 g_1\), \( = g_1^{n_1 m_1} g_2^{n_2 m_2} = (g_1^{n_1})^{m_1} (g_2^{n_2})^{m_2} = 1^{m_1} 1^{m_2} = 1 1 = 1\).
So, \(n := \vert (g_1 g_2) \vert\) is a divisor of \(l\).
Step 3:
Let us suppose that \(n \lt l\).
\(n\) would be a common divisor of \(n_1\) and \(n_2\), because \(gcd (m_1, m_2) = 1\).
So, let \(n_1 = n n'_1\) and \(n_2 = n n'_2\).
\(((g_1 g_2)^n)^{n'_1} = 1^{n'_1} = 1\), but \(((g_1 g_2)^n)^{n'_1} = (g_1 g_2)^{n n'_1} = (g_1 g_2)^{n_1} = g_1^{n_1} g_2^{n_1} = 1 g_2^{n_1} = g_2^{n_1}\).
So, \(g_2^{n_1} = 1\), which is a contradiction against \(\vert (g_2) \vert = n_2\), because \(n_1 \lt n_2\).
So, \(n = l\).
3: Note
When \(n_1 = n_2\), \(\vert (g_1 g_2) \vert\) may not equal \(lcm (n_1, n_2)\): as a counterexample, \(g_1 = g_2\) and \(n_1 = n_2 = 4\), then, \(g_1 g_2 = g_1^2\), and \((g_1 g_2)^2 = (g_1^2)^2 = g_1^4 = 1\), so, \(\vert (g_1 g_2) \vert = 2 \neq lcm (4, 4) = 4\).
