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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Z(Sn) is trivial for n3

Statement

Z(Sn)={1} for n3. In contrast, S0, S1, and S2 are abelian, so their centers are the whole groups.

Facts & Assumptions

Given: The symmetric groups Sn.

[F1]

The center consists of elements commuting with every group element (The center Z(G) of a group).

Proof

technique · contradiction
1.1

For n=0,1, the groups are trivial; S2 has two elements and is cyclic. Hence all three are abelian, as is also consistent with the conjugacy and centralizer descriptions in [F3].

F3algebra
1.2

Let n3 and suppose, for contradiction, that σ1 lies in Z(Sn). Choose a with b=σ(a)a and choose c distinct from a,b.

assume-contrachoose
2.1

For t=(ac), [F2] gives σtσ1=(bσ(c)). This cannot equal (ac) because its support contains b{a,c}.

F2step 1.2
3.1

Thus σ does not commute with t, contradicting [F1] and the assumption that it is central. Therefore only the identity is central for n3, while step 1.1 gives the stated exceptional centers.

F1step 1.1step 2.1discharge-contradiction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 11 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