description/proof of that for group and normal subgroup, if normal subgroup and quotient of group by normal subgroup are p-groups, group is p-group
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader knows a definition of p-group.
- The reader knows a definition of subgroup generated by subset of group.
- The reader admits the proposition that for any group and its any finite-order element, the order power of the element is \(1\) and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
- The reader admits the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any normal subgroup, if the normal subgroup and the quotient of the group by the normal subgroup are p-groups, the group is a p-group.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(G' / G\): \(= \text{ the quotient group }\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
//
Statements:
\(G, G' / G \in \{\text{ the p-groups }\}\)
\(\implies\)
\(G' \in \{\text{ the p-groups }\}\)
//
2: Proof
Whole Strategy: Step 1: take each \(g' \in G'\) and \((G g')\) and suppose that the order of \((G g')\) is \(p^n\), and see that \(g'^{p^n}\) is the 1st such that \(g'^k \in G\) where \(k \in \mathbb{N} \setminus \{0\}\); Step 2: suppose that the order of \((g'^{p^n})\) is \(p^m\), and see that \((g'^{p^n})^{p^m}\) is the 1st such that \(g^k = 1\) where \(k \in \mathbb{N} \setminus \{0\}\).
Step 1:
Let us take each \(g' \in G'\).
Let us take \((G g') \in \{\text{ the subgroups of } G' / G\}\).
Let us suppose that the order of \((G g')\) is \(p^n\): \(G' / G\) is a p-group.
\((G g') = \{1, G g', ..., G g'^{p^n - 1}\}\) where \(G g'^{p^n} = 1\), by the proposition that for any group and its any finite-order element, the order power of the element is \(1\) and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
That means that \(g'^{p^n}\) is the 1st such that \(g'^k \in G\) where \(k \in \mathbb{N} \setminus \{0\}\): \(G g'^k = 1\) means that \(g'^k \in G\).
Step 2:
Let us think of \((g')\).
\((g') = \{1, g', g'^{-1}, g'^2, g'^{-2}, ...\}\), which may have some duplications.
We already know that \(g'^{p^n} \in G\).
Let us suppose that the order of \((g'^{p^n})\) is \(p^m\): \(G\) is a p-group.
\((g'^{p^n}) = \{1, g'^{p^n}, ..., (g'^{p^n})^{p^m - 1}\}\) where \((g'^{p^n})^{p^m} = 1\), by the proposition that for any group and its any finite-order element, the order power of the element is \(1\) and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
That means that \((g'^{p^n})^{p^m}\) is the 1st such that \((g'^{p^n})^k = 1\) where \(k \in \mathbb{N} \setminus \{0\}\).
Let us suppose that \(g'^k = 1\) where \(k \in \mathbb{N} \setminus \{0\}\).
Let \(k = l p^n + j\) where \(l \in \mathbb{N}\) and \(j \in \mathbb{N}\) such that \(0 \le j \lt p^n\).
\(g'^k = g'^{l p^n + j} = g'^{l p^n} g'^j = 1\), which implies that \(j = 0\), because \(g'^j = (g'^{l p^n})^{-1} = g'^{- l p^n} = (g'^{p^n})^{- l} \in G\), but \(g'^{p^n}\) is the 1st such that \(g'^k \in G\) where \(k \in \mathbb{N} \setminus \{0\}\).
So, \(k = l p^n\) where \(l \in \mathbb{N} \setminus \{0\}\).
But \(p^m \le l\), because \(g^k = g^{l p^n} = (g^{p^n})^l\), but \((g'^{p^n})^{p^m}\) is the 1st such that \((g'^{p^n})^k = 1\) where \(k \in \mathbb{N} \setminus \{0\}\).
\((g'^{p^n})^{p^m} = g'^{p^n p^m} = g'^{p^{n + m}} = 1\).
So, \((g') = \{1, g', ..., g'^{p^{n + m} - 1}\}\), by the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.
So, the order of \(g'\) is \(p^{n + m}\), which means that \(G'\) is a p-group.
