description/proof of that for finite \(p\)-group, for natural number smaller than power to which \(p\) is order of group, there is normal subgroup of group whose order is \(p\) to power of natural number
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of \(p\)-group.
- The reader knows a definition of normal subgroup of group.
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader knows a definition of center of group.
- The reader knows a definition of union of set.
- The reader admits the proposition that for any finite group, the group is a \(p\)-group if and only if the order of the group is \(p\) to the power of a natural number.
- The reader admits the proposition that for any \(p\)-group, its center is not trivial.
- The reader admits Lagrange's theorem.
- The reader admits Cauchy's theorem.
- The reader admits the correspondence theorem for group and quotient group.
Target Context
- The reader will have a description and a proof of the proposition that for any finite \(p\)-group, for each natural number smaller than the power to which \(p\) is the order of the group, there is a normal subgroup of the group whose order is \(p\) to the power of the natural number.
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:
\(G\): \(\in \{\text{ the finite } p \text{ -groups }\}\)
//
Statements:
\(\exists n \in \mathbb{N} (\vert G \vert = p^n)\)
\(\land\)
\(\forall k \in \mathbb{N} \text{ such that } 0 \le k \lt n (\exists G_k \in \{\text{ the normal subgroups of } G\} (\vert G_k \vert = p^k))\)
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2: Proof
Whole Strategy: prove it inductively; Step 1: admit that \(\vert G \vert = p^n\); Step 2: prove it for \(k = 0\); Step 3: suppose that there are some \(G_0, ..., G_{l - 1}\) and see that there is a \(G_l\).
Step 1:
For any finite \(p\)-group, there is an \(n \in \mathbb{N}\) such that \(\vert G \vert = p^n\), by the proposition that for any finite group, the group is a \(p\)-group if and only if the order of the group is \(p\) to the power of a natural number, which is not the main point of this proposition but is a preparation.
Step 2:
Let us suppose that \(k = 0\).
\(G_0 := \{1\}\) is the normal subgroup of \(G\) such that \(\vert G_0 \vert = p^0 = 1\).
Step 3:
Step 3 Strategy: Step 3-1: suppose that there are \(G_0, ..., G_{l - 1}\); Step 3-2: take \(G / G_{l - 1}\) and see that it is a \(p\)-group; Step 3-3: take \(Z (G / G_{l - 1})\) and see that it is a \(p\)-group; Step 3-4: take a subgroup of \(Z (G / G_{l - 1})\), \(N\), such that \(\vert N \vert = p\) and see that \(N\) is a normal subgroup of \(G / G_{l - 1}\); Step 3-5: see that \(G_l := \cup N\) is a normal subgroup of \(G\) such that \(\vert G_l \vert = p^l\).
Step 3-1:
Let us suppose that there are \(G_0, ..., G_{l - 1}\).
Step 3-2:
Let us think of the quotient, \(G / G_{l - 1}\), which is valid because \(G_{l - 1}\) is a normal subgroup of \(G\).
\(\vert G / G_{l - 1} \vert = p^n / p^{l - 1} = p^{n - (l - 1)}\), by the definition of quotient group of group by normal subgroup.
So, \(G / G_{l - 1}\) is a \(p\)-group, by the proposition that for any finite group, the group is a \(p\)-group if and only if the order of the group is \(p\) to the power of a natural number.
Step 3-3:
Let us think of the center of \(G / G_{l - 1}\), \(Z (G / G_{l - 1})\).
\(Z (G / G_{l - 1})\) is a normal subgroup of \(G / G_{l - 1}\).
\(1 \lt \vert Z (G / G_{l - 1}) \vert\), by the proposition that for any \(p\)-group, its center is not trivial, which means that \(\vert Z (G / G_{l - 1}) \vert = p^m\) for an \(m \in \mathbb{N} \setminus \{0\}\), by Lagrange's theorem, which means that \(Z (G / G_{l - 1})\) is a \(p\)-group by the proposition that for any finite group, the group is a \(p\)-group if and only if the order of the group is \(p\) to the power of a natural number.
Step 3-4:
There is a subgroup of \(Z (G / G_{l - 1})\), \(N\), such that \(\vert N \vert = p\), by Cauchy's theorem.
\(N\) is a normal subgroup of \(G / G_{l - 1}\), because \(N\) is a subgroup of the center: for each \(g \in G / G_{l - 1}\), \(g N g^{-1} = N\), because for each \(g' \in N\), \(g g' g^{-1} = g'\), because \(g' \in Z (G / G_{l - 1})\).
Step 3-5:
Then, \(\cup N\) is a normal subgroup of \(G\), by the correspondence theorem for group and quotient group.
\(\vert \cup N \vert = \vert G_{l - 1} \vert \vert N \vert = p^{l - 1} p = p^l\).
So, we can take \(G_l := \cup N\).
