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$\operatorname{Sym}(\{0,1,2\})\cong\langle s,t\mid s^2,\ t^2,\ (st)^3\rangle$

Example

With rightmost-first composition and transpositions $s=(0\,1)$ and $t=(1\,2)$ ,

$\operatorname{Sym}(\{0,1,2\})\cong\langle s,t\mid s^2,\ t^2,\ (st)^3\rangle.$

Facts & Assumptions

Given: The set $A=\{0,1,2\}$ , the permutations $s=(0\,1)$ and $t=(1\,2)$ , and the presentation $P=\langle s,t\mid s^2,t^2,(st)^3\rangle$ .

[L1]

If a finite set $A$ has $n$ elements, then the set of its bijections has cardinality $n!$ (A finite set $A$ with $\lvert A\rvert = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

[L2]

A map of generators that sends every relator to the identity extends uniquely to a homomorphism from the presented group (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

Verification

technique · constructive

1.1

Direct permutation computation gives $s^2=t^2=\operatorname{id}$ and $(st)^3=\operatorname{id}$ , so [L2] constructs a homomorphism $\pi:P\to\operatorname{Sym}(A)$ .

L2construct

1.2

After cancelling $s^2$ and $t^2$ , every word alternates. The relation $(st)^3=e$ gives $stst=ts$ and hence $tst=sts$ after multiplying on the left by $s$ ; replacing the first three letters of any alternating word of length at least four by the other side creates an adjacent equal pair and shortens the word.

given

1.3

Their images are respectively $\operatorname{id},(0\,1),(1\,2),(0\,1\,2),(0\,2\,1),(0\,2)$ , so they are distinct; [L1] gives $|\operatorname{Sym}(A)|=3!=6$ , and these images exhaust it.

L1given

2.1

Repeating step 1.2 leaves one of $e,s,t,st,ts,sts$ , since the two alternating words of length three are equal.

step 1.2

3.1

Step 2.1 gives at most six elements in $P$ , while step 1.3 gives six distinct images under $\pi$ ; hence $\pi$ is bijective and is the claimed isomorphism.

step 1.1step 1.3step 2.1discharge-construct∎

Depends on

Used by

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Sources

Encyclopedia of Mathematics, Presentation (standard reference, not scraped)

Sym⁡({0,1,2})≅⟨s,t∣s2, t2, (st)3⟩\operatorname{Sym}(\{0,1,2\})\cong\langle s,t\mid s^2,\ t^2,\ (st)^3\rangleSym({0,1,2})≅⟨s,t∣s2, t2, (st)3⟩