Help with proof: If a is in S_n and commutes with every b in S_n, then a=(1)

**didact273** · Feb 11th 2009, 08:56 PM

Prove that for n>=3, if a is in S_n and commutes with every b in S_n,
then a=(1)

**math2009** · Feb 11th 2009, 09:07 PM

$ab=ba$ ?

Are $a,b$ matrix ?

**didact273** · Feb 11th 2009, 09:12 PM

Originally Posted by math2009

$ab=ba$ ?

Are $a,b$ matrix ?

No a and b are permutations in S_n.

S_n being the symmetric group on n letters.

**NonCommAlg** · Feb 11th 2009, 10:00 PM

Originally Posted by didact273

Prove that for n>=3, if a is in S_n and commutes with every b in S_n, then a=(1)

let $(1) \neq a \in S_n.$ so there exist $1 \leq i \neq j \leq n$ with $a(i)=j.$ since $n \geq 3,$ there exists $1 \leq k \leq n$ such that $k \notin \{i,j\}.$

now let $b=(j \ \ k).$ then $(ab)(i)=a(i)=j \neq k =b(j)=(ba)(i).$ thus $ab \neq ba. \ \Box$

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