description/proof of that for 4-symmetric group, 4-alternating group is only subgroup with order 12
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of n-symmetric group.
- The reader knows a definition of n-alternating group.
- The reader admits the proposition that for any sequence of any finite elements, the set of the permutations has the same number of the even permutations and the odd permutations.
Target Context
- The reader will have a description and a proof of the proposition that for the 4-symmetric group, the 4-alternating group is the only subgroup with order 12.
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:
\(S_4\): \(= \text{ the } 4 \text{ -symmetric group }\)
\(A_4\): \(= \text{ the } 4 \text{ -alternating group }\)
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Statements:
\(\{\text{ the subgroups of } S_4 \text{ with order 12 }\} = \{A_4\}\)
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2: Proof
Whole Strategy: Step 1: see that \(A_4\) is a subgroup of \(S_4\) with order 12; Step 2: see that \(S_4\) consists of \(1\), 6 2-cycles, 8 3-cycles, 6 4-cycles, and 3 products of disjoint 2-cycles; Step 3: see that any subgroup of order 12 cannot have 3 4-cycles; Step 4: see that any subgroup of order 12 cannot have only 2 4-cycles; Step 5: supposing that a subgroup of order 12 has no 4-cycle, see that the subgroup cannot have any 2-cycle.
Step 1:
\(A_4\) is a subgroup of \(S_4\): see Note for the definition of n-alternating group.
\(\vert S_4 \vert = 4! = 24\). \(\vert A_4 \vert = \vert S_4 \vert / 2 = 24 / 2 = 12\), by the proposition that for any sequence of any finite elements, the set of the permutations has the same number of the even permutations and the odd permutations.
Step 2:
\(S_4\) consists of \(1\), 6 2-cycles, 8 3-cycles, 6 4-cycles, and 3 products of disjoint 2-cycles: \(6 = 4 * 3 / 2\); \(8 = 4 * 3 * 2 / 3\); \(6 = 4 * 3 * 2 * 1 / 4\); \(3 = 4 * 3 / 2 / 2\): for counting the elements of the type \((a, b) (c, d)\), \((a, b)\) s have \(4 * 3 / 2\) patterns (for each \((a, b)\), \((c, d)\) is uniquely determined), but \((a, b) (c, d)\) and \((c, d) (a, b)\) are being doubly counted, so, the last \( / 2\) is due.
Step 3:
Hereafter, the memorandum on some multiplications of cycles on any symmetric group is used.
Let us suppose that there was a subgroup with order 12 that (the subgroup) was not \(A_4\).
That means that the subgroup contained at least 1 odd permutation.
The odd permutations are the 6 2-cycles and the 6 4-cycles: for example, \((1, 2, 3, 4) = (2, 3) (3, 4) (1, 4)\).
Let us see that the subgroup cannot have 3 4-cycles.
Let us suppose that the subgroup had some 3 4-cycles and find a contradiction.
Let \((1, 2, 3, 4)\) be the 1st 4-cycle in the subgroup: we do not need to consider any other case like \((1, 2, 4, 3)\), because that would be just a renaming of the elements of \(\{1, 2, 3, 4\}\): what matters is the relation between the 1st 4-cycle and the 3rd 4-cycle (the 2nd 4-cycle is automatically determined as is seen below), not what the 1st 4-cycle is.
The subgroup had \(\{1, (1, 2, 3, 4), (1, 2, 3, 4)^2 = (1, 3) (2, 4), (1, 2, 3, 4)^3 = (1, 4, 3, 2)\}\).
We see that the subgroup inevitably had the 2nd 4-cycle, \((1, 4, 3, 2)\).
We are going to see that the subgroup could not have any 3rd 4-cycle.
The only cases to be consider are that the 3rd 4-cycle was \((1, 2, 4, 3)\) or \((1, 3, 2, 4)\): \((1, 3, 4, 2)\) are \((1, 4, 2, 3)\) contained in the \((1, 2, 4, 3)\) and \((1, 3, 2, 4)\) cases, respectively: \((1, 3, 4, 2)^3 = (1, 2, 4, 3)\) and \((1, 4, 2, 3)^3 = (1, 3, 2, 4)\).
Let \((1, 2, 4, 3)\) be the 3rd 4-cycle.
The subgroup had \(\{1, (1, 2, 4, 3), (1, 2, 4, 3)^2 = (1, 4) (2, 3), (1, 2, 4, 3)^3 = (1, 3, 4, 2)\}\).
\((1, 2, 3, 4) (1, 2, 4, 3) = (1, 3, 2)\).
So, the subgroup had \(\{1, (1, 3, 2), (1, 3, 2)^2 = (1, 2, 3)\}\).
\((1, 2, 4, 3) (1, 2, 3, 4) = (1, 4, 2)\).
So, the subgroup had \(\{1, (1, 4, 2), (1, 4, 2)^2 = (1, 2, 4)\}\).
\((1, 4, 2) (1, 3) (2, 4) = (1, 3, 4)\).
So, the subgroup had \(\{1, (1, 3, 4), (1, 3, 4)^2 = (1, 4, 3)\}\).
So, the subgroup had more than \(12\) elements.
Let \((1, 3, 2, 4)\) be the 3rd 4-cycle.
The subgroup had \(\{1, (1, 3, 2, 4), (1, 3, 2, 4)^2 = (1, 2) (3, 4), (1, 3, 2, 4)^3 = (1, 4, 2, 3)\}\).
\((1, 2, 3, 4) (1, 3, 2, 4) = (1, 4, 2)\).
So, the subgroup had \(\{1, (1, 4, 2), (1, 4, 2)^2 = (1, 2, 4)\}\).
\((1, 3, 2, 4) (1, 2, 3, 4) = (1, 4, 3)\).
So, the subgroup had \(\{1, (1, 4, 3), (1, 4, 3)^2 = (1, 3, 4)\}\).
\((1, 4, 3) (1, 3) (2, 4) = (2, 3, 4)\).
So, the subgroup had \(\{1, (2, 3, 4), (2, 3, 4)^2 = (2, 4, 3)\}\).
So, the subgroup had more than \(12\) elements.
So, the subgroup cannot have more than 2 4-cycles: the only possibilities are that the subgroup has no 4-cycle or only 2 4-cycles.
Step 4:
Let us see that the subgroup cannot have only 2 4-cycles.
Let us suppose that the subgroup had only 2 4-cycles.
Let \((1, 2, 3, 4)\) be the 1st 4-cycle in the subgroup: we do not need to consider any other case like \((1, 2, 4, 3)\), because that would be just a renaming of the elements of \(\{1, 2, 3, 4\}\).
The subgroup had \(\{1, (1, 2, 3, 4), (1, 2, 3, 4)^2 = (1, 3) (2, 4), (1, 2, 3, 4)^3 = (1, 4, 3, 2)\}\).
Step 4 Strategy: see what cannot be in the subgroup and see that only the remained elements cannot generate the subgroup.
\((1, 2)\) cannot be in the subgroup, because \((1, 2) (1, 3) (2, 4) = (1, 3, 2, 4)\), which cannot be in the subgroup by the supposition of this step.
\((3, 4)\) cannot be in the subgroup, because \((3, 4) (1, 3) (2, 4) = (1, 4, 2, 3)\), which cannot be in the subgroup by the supposition of this step.
\((1, 2, 3)\) cannot be in the subgroup, because \((1, 2, 3) (1, 2, 3, 4) = (1, 3, 4, 2)\), which cannot be in the subgroup by the supposition of this step.
\((1, 2, 4)\) cannot be in the subgroup, because \((1, 2, 4) (1, 2, 3, 4) = (1, 4, 2, 3)\), which cannot be in the subgroup by the supposition of this step.
\((1, 3, 2)\) cannot be in the subgroup, because \((1, 3, 2)^2 = (1, 2, 3)\), which is already known to not be in the subgroup.
\((1, 4, 3)\) cannot be in the subgroup, because \((1, 4, 3) (1, 2, 3, 4) = (1, 2)\), which is already known to not be in the subgroup.
\((2, 3, 4)\) cannot be in the subgroup, because \((2, 3, 4) (1, 2, 3, 4) = (1, 3, 2, 4)\), which cannot be in the subgroup by the supposition of this step.
\((2, 4, 3)\) cannot be in the subgroup, because \((2, 4, 3)^2 = (2, 3, 4)\), which is already known to not be in the subgroup.
So, 2 2-cycles and 6 3-cycles cannot be in the subgroup.
As also the other 4 4-cycles are not in the subgroup, \(2 + 6 + 4 = 12\) elements are already excluded, and all the other elements need to be in the subgroup.
So, \((1, 3, 4)\) and \((2, 4)\) need to be in the subgroup, but \((1, 3, 4) (2, 4) = (1, 3, 4, 2)\), which cannot be in the subgroup by the supposition of this step.
So, the subgroup does not exist.
Step 5:
Let us suppose that the subgroup has no 4-cycle.
Let us suppose that \((1, 2)\) was in the subgroup: we do not need to consider any other case like \((3, 4)\), because that would be just a renaming of the elements of \(\{1, 2, 3, 4\}\).
Let us see what could not be in the subgroup.
\((1, 3, 4)\) was not in the subgroup, because \((1, 2) (1, 3, 4) = (1, 3, 4, 2)\), which could not be in the subgroup by the supposition of this step.
\((1, 4, 3)\) was not in the subgroup, because \((1, 2) (1, 4, 3) = (1, 4, 3, 2)\), which could not be in the subgroup by the supposition of this step.
\((2, 3, 4)\) was not in the subgroup, because \((1, 2) (2, 3, 4) = (1, 2, 3, 4)\), which could not be in the subgroup by the supposition of this step.
\((2, 4, 3)\) was not in the subgroup, because \((1, 2) (2, 4, 3) = (1, 2, 4, 3)\), which could not be in the subgroup by the supposition of this step.
\((1, 3) (2, 4)\) was not in the subgroup, because \((1, 2) (1, 3) (2, 4) = (1, 3, 2, 4)\), which could not be in the subgroup by the supposition of this step.
\((1, 4) (2, 3)\) was not in the subgroup, because \((1, 2) (1, 4) (2, 3) = (1, 4, 2, 3)\), which could not be in the subgroup by the supposition of this step.
So, the 4 3-cycles and the 2 products of disjoint 2-cycles were excluded.
As also all the 6 4-cycles were excluded, \(4 + 2 + 6 = 12\) elements were already excluded, and all the other elements need to be in the subgroup.
So, \((1, 3)\) and \((2, 4)\) need to be in the subgroup, but \((1, 3) (2, 4)\) could not be in the subgroup.
So, no 2-cycle cannot be in the subgroup.
That means that no odd permutation can be in the subgroup.
So, \(A_4\) is the only 12-ordered subgroup of \(S_4\).
