description/proof of that Abelian group is simple group iff its order is prime number
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of Abelian group.
- The reader knows a definition of simple group.
- The reader admits Lagrange's theorem.
Target Context
- The reader will have a description and a proof of the proposition that any Abelian group is a simple group if and only if its order is a prime 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:
\(G\): \(\in \{\text{ the Abelian groups }\}\)
//
Statements:
\(G \in \{\text{ the simple groups }\}\)
\(\iff\)
\(\vert G \vert = p\) where \(p \in \{\text{ the prime numbers }\}\)
//
2: Natural Language Description
Any Abelian group is a simple group if and only if the order, \(\vert G \vert\) is any prime number, \(p\).
3: Proof
Whole Strategy: Step 1: suppose that \(G\) is a simple group, take any element, \(g \in G \setminus \{1\}\), and the subgroup generated by \(g\), \((g)\), and see that \((g) = G\); Step 2: when \(g^2 = 1\), see that \(\vert G \vert = 2\), otherwise, take \((g^2)\) and see that \((g^2) = G = (g)\); Step 3: see that Step 1 and Step 2 imply that \(\vert G \vert\) is finite; Step 4: let \(\vert G \vert = p k\) for a prime number, \(p\), and see that \(k = 1\); Step 5: suppose that \(\vert G \vert = p\) for a prime number, \(p\), and see that its subgroups are only \(\{1\}\) and \(G\).
Step 1:
Let us suppose that \(G\) is a simple group.
Let us take any element, \(g \in G \setminus \{1\}\), and the subgroup generated by \(g\), \((g) = \{g' \in G \vert g' = g^z, z \in \mathbb{Z}\}\).
\((g)\) is indeed a subgroup: \(g^z g^{z'} = g^{z + z'} \in (g)\); \(g^0 = 1 \in (g)\); for each \(g^z \in (g)\), \(g^z g^{-z} = g^{-z} g^z = g^0 = 1\).
\((g)\) is a normal subgroup, because each subgroup of any Abelian group is a normal subgroup.
So, \((g) = G\), because \(G\) is simple (obviously, \(G \neq \{1\}\)).
Step 2:
When \(g^2 = 1\), \(G = \{1, g\}\), and \(\vert G \vert = 2\), which is a prime number.
Let us suppose otherwise hereafter.
Let us take \((g^2)\), which is a normal subgroup of \(G\), as before, because we just have took \(g^2\) instead of \(g\).
\((g^2) = G\), as before.
Step 3:
Let us see that \(G = (g) = (g^2)\) implies that \(\vert G \vert\) is finite.
There is a \(z'\) such that \(g = g^{2 z'}\).
\(1 = g g^{-1} = g^{2 z'} g^{-1} = g^{2 z' - 1}\). \(2 z' - 1 \neq 0\), because it is odd. When \(0 \lt 2 z' - 1\), let \(n := 2 z' - 1\); otherwise, \(0 \lt - (2 z' - 1)\), and let \(n := - (2 z' - 1)\). Anyway, \(g^n = 1\), because \(1 = 1^{-1} = (g^{2 z' - 1})^{-1} = g^{- (2 z' - 1)}\). \(n \neq 1\), because \(g^1 = g \neq 1\). \(g^{-1} = g^{n - 1}\), because \(g g^{n - 1} = g^{n - 1} g = g^n = 1\).
So, each element of \((g)\) equals one of \(\{1, g, ..., g^{n - 1}\}\) (there may be some duplications in the set), because for each \(z \in \mathbb{Z}\), \(z = n k + l\) where \(0 \le l \lt n\) and \(g^z = g^{n k + l} = g^{n k} g^l = (g^n)^k g^l = 1^k g^l = g^l\).
So, \(\vert G \vert = \vert (g) \vert\) is finite.
Step 4:
Let us take the prime factorization of \(\vert G \vert\). As \(1 \lt \vert G \vert\), there is at least 1 prime number, \(p\), such that \(\vert G \vert = p k\) where \(k \in \mathbb{N} \setminus \{0\}\).
As \(G = (g)\), \(G = \{1, g, ..., g^{p k - 1}\}\) where \(g^{p k} = 1\).
Let us take \((g^p)\). As \(g^{p k} = 1\), \((g^p) = \{1, g^p, ..., g^{p (k - 1)}\}\).
As \((g^p)\) is a normal subgroup as before and \(G\) is simple, \((g^p)\) is \(\{1\}\) or \(G\). But it cannot be \(G\), because at least \(g, ..., g^{p - 1}\) are missing there. So, \((g^p) = \{1\}\). That means that \(k = 1\).
So, \(\vert G \vert = p\).
Step 5:
Let us suppose that \(\vert G \vert = p\) for a prime number, \(p\).
By Lagrange's theorem, each of the only possible subgroups of \(G\) has order \(1\) or \(p\). So, the subgroups of \(G\) are only \(\{1\}\) and \(G\). The normal subgroups of \(G\) are \(\{1\}\) and \(G\), even more.
So, \(G\) is simple.
