952: For 4-Symmetric Group, 4-Alternating Group Is Only Subgroup with Order 12

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description/proof of that for 4-symmetric group, 4-alternating group is only subgroup with order 12

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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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.

$|S_4|=4!=24$. $|A_4|=|S_4|/2=24/2=12$, 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: $6=4\ast 3/2$; $8=4\ast 3\ast 2/3$; $6=4\ast 3\ast 2\ast 1/4$; $3=4\ast 3/2/2$: for counting the elements of the type $(a,b)(c,d)$, $(a,b)$ s have $4\ast 3/2$ 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 $S_4$.

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References

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