description/proof of that 6-elements group cannot have 2 3-elements subgroups that share only identity
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 group.
Target Context
- The reader will have a description and a proof of the proposition that any 6-elements group cannot have any 2 3-elements subgroups that share only the identity.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(G_1\): \(\in \{\text{ the subgroups of } G\}\)
\(G_2\): \(\in \{\text{ the subgroups of } G\}\)
//
Statements:
(
\(\vert G \vert = 6\)
\(\land\)
\(\vert G_1 \vert = 3\)
\(\land\)
\(G_1 \cap G_2 = \{1\}\)
)
\(\implies\)
\(\vert G_2 \vert \neq 3\)
//
2: Natural Language Description
Any 6-elements group, \(G\), cannot have any 2 3-elements subgroups, \(G_1, G_2\), such that \(G_1 \cap G_2 = \{1\}\).
3: Proof
Let us suppose that there were \(G_1 = \{1, a_1, a_2\}\) and \(G_2 = \{1, b_1, b_2\}\) such that \(G_1 \cap G_2 = \{1\}\).
\(\{1, a_1, a_2, b_1, b_2\}\) would be distinct, and there would be the leftover, \(c \in G\).
What would be \(b_1 a_1\)? It could not be \(1\), because \(b_1 a_1 = 1\) would imply \(b_1 = {a_1}^{-1} \in G_1 \cap G_2 = \{1\}\), which would imply \(b_1 = 1\), a contradiction. It could not be \(a_j\), because \(b_1 a_1 = a_j\) would imply \(b_1 = a_j {a_1}^{-1} \in G_1 \cap G_2 = \{1\}\), which would imply \(b_1 = 1\), a contradiction. It could not be \(b_j\), because \(b_1 a_1 = b_j\) would imply \(a_1 = {b_1}^{-1} b_j \in G_1 \cap G_2 = \{1\}\), which would imply \(a_1 = 1\), a contradiction. So, \(b_1 a_1 = c\). But likewise, \(b_1 a_2 = c\), then, \(a_1 = {b_1}^{-1} c = a_2\), a contradiction. So, \(b_1 a_1\) cannot have any valid value.
So, the supposition is false.
