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For $n \geq 5$ , the only proper nontrivial normal subgroup of $S_{n}$ is $A_{n}$

Statement

For $n \geq 5$ , the normal subgroups of $S_{n}$ are ${1}$ , $A_{n}$ , and $S_{n}$ . Thus $A_{n}$ is the only proper nontrivial one.

Facts & Assumptions

Given: $n \geq 5$ and $N ⊴ S_{n}$ .

[F1]

$A_{n} ⊴ S_{n}$ and $2 ∣ A_{n} ∣ = n!$ for $n \geq 2$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ); with $∣ S_{n} ∣ = n!$ and $∣ S_{n} ∣ = [S_{n} : A_{n}] ∣ A_{n} ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ) this gives $[S_{n} : A_{n}] = 2$ , so $S_{n}$ is the union of $A_{n}$ and one other coset.

[F2]

$A_{n}$ is simple for $n \geq 5$ ( $A_{n}$ is simple for every $n \geq 5$ ).

[F3]

Sign is a homomorphism (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ) with kernel $A_{n}$ (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[F4]

$Z (S_{n}) = {1}$ for $n \geq 3$ ( $Z (S_{n})$ is trivial for $n \geq 3$ ).

[F5]

A subgroup is normal exactly when it is invariant under conjugation (Normal subgroup: invariance under conjugation).

Proof

technique · cases

1.1

The intersection $N \cap A_{n}$ is normal in $A_{n}$ by [F5], so [F2] makes it either ${1}$ or $A_{n}$ .

F2F5algebra

1.2

Suppose $N \cap A_{n} = A_{n}$ .

assume-case full

1.3

Suppose $N \cap A_{n} = {1}$ .

assume-case trivial

2.1

In the case of step 1.2, $A_{n} \subseteq N$ . Since [F1] gives only two cosets, $N = A_{n}$ or $N = S_{n}$ .

F1step 1.2algebra

2.2

In the case of step 1.3, the restriction of sign to $N$ is injective by [F3], so $∣ N ∣ \leq 2$ .

F3step 1.3algebra

3.1

If the subgroup in step 2.2 were nontrivial, it would have a unique nonidentity element $t$ . Normality [F5] would make every conjugate of $t$ the same unique element, so $t \in Z (S_{n})$ , contradicting [F4]. Thus this case gives $N = {1}$ .

F4F5step 2.2

4.1

The alternatives in step 1.1 are exhaustive, and steps 2.1 and 3.1 give exactly $N \in {{1}, A_{n}, S_{n}}$ .

step 1.1step 2.1step 3.1cases-exhaustive∎

Depends on

Used by

A nonsurjective homomorphism need not carry the Frattini subgroup into the target Frattini subgroup Counterexample

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory (standard reference, not scraped)

For n≥5n≥5, the only proper nontrivial normal subgroup of SnSn​ is AnAn​