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$Z (S_{n})$ is trivial for $n \geq 3$

Statement

$Z (S_{n}) = {1}$ for $n \geq 3$ . In contrast, $S_{0}$ , $S_{1}$ , and $S_{2}$ are abelian, so their centers are the whole groups.

Facts & Assumptions

Given: The symmetric groups $S_{n}$ .

[F1]

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

[F2]

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

[F3]

Conjugate permutations are classified by cycle type (Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type), and a centralizer has the cycle-type cardinality formula (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

Proof

technique · contradiction

1.1

For $n = 0, 1$ , the groups are trivial; $S_{2}$ 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 $n \geq 3$ and suppose, for contradiction, that $σ \neq 1$ lies in $Z (S_{n})$ . Choose $a$ with $b = σ (a) \neq a$ and choose $c$ distinct from $a, b$ .

assume-contrachoose

2.1

For $t = (a c)$ , [F2] gives $σ t σ^{- 1} = (b σ (c))$ . This cannot equal $(a c)$ because its support contains $b \notin {a, c}$ .

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 $n \geq 3$ , while step 1.1 gives the stated exceptional centers.

F1step 1.1step 2.1discharge-contradiction∎

Depends on

Used by

For n≥5, the only proper nontrivial normal subgroup of Sₙ is Aₙ Corollary

Dependency tree · two levels

11 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

K. Conrad, Conjugacy Classes (standard reference, not scraped)

Z(Sn)Z(Sn​) is trivial for n≥3n≥3