description/proof of that for finite cyclic group and its prime factor of order, there is at most 1 subgroup of factor order
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.
Target Context
- The reader will have a description and a proof of the proposition that for any finite cyclic group and its any prime factor of the order of the group, there is at most 1 subgroup of the factor order.
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:
\(p\): \(\in \{\text{ the prime numbers }\}\)
\(G\): \(\in \{\text{ the cyclic groups }\}\) with order \(p n\), \(= \{g , ..., g^{p n} = 1\}\)
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Statements:
\(\vert \{\text{ the } p \text{ -ordered subgroups of } G\} \vert \le 1\)
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2: Note
By the proposition that for any finite cyclic group and any prime factor of the order of the group, a cyclic subgroup of the factor order can be extracted in a certain way, there is the cyclic subgroup of the \(p\) order mentioned in the proposition, and in fact, that is the only \(p\)-ordered subgroup of \(G\), by this proposition.
3: Proof
Whole Strategy: Step 1: see that any \(p\)-ordered subgroup is a cyclic group generated by a \(g^r\) where \(1 \le r \lt p n\); Step 2: see the condition that \(r\) needs to satisfy and that there are only \(p - 1\) possibilities; Step 3: conclude the proposition.
Step 1:
Any \(p\)-ordered subgroup is a cyclic group, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.
So, it is generated by an element of \(G\), and let the element be \(g^r\) where \(1 \le r \lt p n\).
Step 2:
\(r\) needs to satisfy \((g^r)^p = 1\).
\((g^r)^p = g^{r p}\). \(r p = p n m\) for an \(m \in \mathbb{N} \setminus \{0\}\). So, \(r = n m\).
As \(1 \le r \lt p n\), the only possibilities are \(m = 1, ..., p - 1\).
So, only those \(p - 1\) elements can possibly generate any \(p\)-ordered subgroup.
Step 3:
If there were 2 different \(p\)-ordered subgroups, the 2 subgroups would share only \(1\), by the proposition that any 2 different prime-number-ordered subgroups share only 1. So, the other different \(2 (p - 1)\) elements would generate a \(p\)-ordered subgroup, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, which is a contradiction against the fact that only those \(p - 1\) elements can possibly generate any \(p\)-ordered subgroup.
So, there is at most 1 \(p\)-ordered subgroup.
