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$V_{4} = {1, (12) (34), (13) (24), (14) (23)}$ is a proper nontrivial normal subgroup of $A_{4}$

Example

In $A_{4}$ , set $V_{4} = {1, (12) (34), (13) (24), (14) (23)} .$ Then $V_{4}$ is a proper nontrivial normal subgroup.

Facts & Assumptions

Given: The displayed subset $V_{4} \subseteq A_{4}$ .

[F1]

$A_{4}$ is the kernel of sign (The alternating group $A_{n} = \ker (sgn)$ of even permutations), and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are included as $1$ -cycles (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[F2]

Conjugation relabels every cycle entry (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F3]

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

[F4]

$∣ A_{4} ∣ = 4! / 2 = 12$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

Verification

technique · direct

1.1

Every displayed double transposition has two cycles and hence sign $(- 1)^{4 - 2} = + 1$ by [F1]. The product of two distinct nonidentity displayed elements is the third, and each is its own inverse; hence $V_{4}$ is a subgroup of $A_{4}$ .

F1algebra

1.2

By [F2], conjugation by any permutation relabels a double transposition to another double transposition. Thus $V_{4}$ is invariant under $A_{4}$ -conjugation, and [F3] makes it normal.

F2F3

2.1

Its order is $4$ , strictly between $1$ and $12 = ∣ A_{4} ∣$ from [F4], so it is nontrivial and proper.

F4algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources

T. Judson, Abstract Algebra: Theory and Applications, Simplicity of $A_n$ (standard reference, not scraped)

V4={1,(12)(34),(13)(24),(14)(23)}V4​={1,(12)(34),(13)(24),(14)(23)} is a proper nontrivial normal subgroup of A4A4​