description/proof of that 2 different prime-number-ordered subgroups share only 1
Topics
About: group
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that any 2 different prime-number-ordered subgroups share only 1.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
\(q\): \(\in \{\text{ the prime numbers }\}\)
\(G_p\): \(\in \{\text{ the } p \text{ -ordered subgroups of } G\}\)
\(G_q\): \(\in \{\text{ the } q \text{ -ordered subgroups of } G\}\)
//
Statements:
\(G_p \neq G_q\)
\(\implies\)
\(G_p \cap G_q = \{1\}\)
//
\(p = q\) is not excluded.
2: Proof
Whole Strategy: Step 1: suppose that there was a \(g \in (G_p \cap G_q) \setminus \{1\}\); Step 2: see that \(G_p = \langle g \rangle = G_q\), a contradiction.
Step 1:
Let us suppose that there was a \(g \in (G_p \cap G_q) \setminus \{1\}\).
Step 2:
\(\langle g \rangle = G_p\), by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.
\(\langle g \rangle = G_q\), likewise.
So, \(G_p = \langle g \rangle = G_q\), a contradiction.
So, the supposition that there was a \(g \in (G_p \cap G_q) \setminus \{1\}\) is wrong, and \((G_p \cap G_q) \setminus \{1\} = \emptyset\), which means that \(G_p \cap G_q = \{1\}\).
