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The subgroup orders in $Sym ({1, 2, 3})$ are $1, 2, 3$ and $6$

Example

Put $S_{3} = Sym ({1, 2, 3})$ . Its subgroup orders are exactly $1, 2, 3,$ and $6$ .

Facts & Assumptions

Given: The symmetric group $S_{3} = Sym ({1, 2, 3})$ under composition.

[F1]

$S_{3}$ is the group of bijections of ${1, 2, 3}$ , with cycle notation and composition acting rightmost first (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[F2]

The generated set $⟨ g ⟩$ is a subgroup; a nonempty subset closed under products and inverses is a subgroup (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, Subgroup, One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[L1]

The order of a subgroup of a finite group divides the order of the group (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Verification

technique · direct

1.1

The six bijections are $e, (12), (13), (23), (123), (132)$ , so $∣ S_{3} ∣ = 6$ .

1.2

The subgroups ${e}$ , $⟨ (12) ⟩ = {e, (12)}$ , $⟨ (123) ⟩ = {e, (123), (132)}$ and $S_{3}$ have orders $1, 2, 3,$ and $6$ , respectively.

F1F2

2.1

If $H \leq S_{3}$ , then [L1] makes $∣ H ∣$ a positive divisor of $6$ , hence $∣ H ∣ \in {1, 2, 3, 6}$ .

step 1.1L1F2

3.1

Step 1.2 realizes every value allowed by step 2.1, so these are exactly the subgroup orders.

step 1.2step 2.1∎

Depends on

Used by

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Dependency tree · two levels

34 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

Thomas W. Judson, Abstract Algebra: Theory and Applications, Cosets and Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.2: Lagrange's Theorem (standard reference, not scraped)

The subgroup orders in Sym⁡({1,2,3})Sym({1,2,3}) are 1,2,31,2,3 and 66