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

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

$\displaystyle ab=ba$ ?

Are $\displaystyle a,b$ matrix ?

Originally Posted by math2009

$\displaystyle ab=ba$ ?

Are $\displaystyle a,b$ matrix ?

No a and b are permutations in S_n.

S_n being the symmetric group on n letters.

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

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