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

Example

Put $S_3=\operatorname{Sym}(\{1,2,3\})$ . Its subgroup orders are exactly $1,2,3,$ and $6$ .

Facts & Assumptions

Given: The symmetric group $S_3=\operatorname{Sym}(\{1,2,3\})$ 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 $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[F2]

The generated set $\langle g\rangle$ is a subgroup; a nonempty subset closed under products and inverses is a subgroup (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups, Subgroup, One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ 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 $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ 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 $\{e\}$ , $\langle(12)\rangle=\{e,(12)\}$ , $\langle(123)\rangle=\{e,(123),(132)\}$ and $S_3$ have orders $1,2,3,$ and $6$ , respectively.

F1F2

2.1

If $H\le S_3$ , 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

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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})\operatorname{Sym}(\{1,2,3\})Sym({1,2,3}) are 1,2,31,2,31,2,3 and 666