Question #223550

If the pair of cycles a=(a1,a2,.......am) and b=(b1,b2,.......bm) have no entries in common .show that ab =ba.


1
Expert's answer
2021-08-05T17:09:21-0400

Suppose α and β are permutations ofS=a1,a2,...,am,b1,b2,...,bn,c1,c2,...,ckwhere the cs are left fixed by both α and βSuppose \space \alpha \space and\space \beta\space are \space permutations \space of\\ S= a_{1}, a_{ 2}, . . . , a_{m}, b_{1}, b_{2}, . . . , b_{n}, c_{1}, c_{2}, . . . , c_{k}\\where \space the\space c’s \space are \space left \space fixed \space by \space both \space \alpha \space and \space \beta

[To show αβ(x)=βα(x) x  S][To \space show \space \alpha \beta(x) = \beta\alpha(x)\forall \space x \space \in \space S ]

ifx=ai for some i,since βfixes all a elements,if x = a_{i} \space for \space some \space i, since\space \beta fixes\space all \space a \space elements,

(αβ)(ai)=α(β(ai))=α(ai)=ai+1(with am+1=a1)and(\alpha\beta)(a_{i})=\alpha(\beta(a_{i}))=\alpha(a_{i})=a _{i+1}(with\space a_{m+1}=a_1)and

(βα)(ai)=β(α(ai))=β(ai)=ai+1,so αβ=βα on the a elements.(\beta\alpha)(a_{i})=\beta(\alpha(a_{i}))=\beta(a_{i})=a _{i+1},\\so\space \alpha\beta=\beta\alpha \space on\space the\space a \space elements.

A similar argument shows αβ=βα on the b elements.A \space similar \space argument\space shows\space \alpha\beta = \beta\alpha\space on\space the\space b\space elements.

Since α and β both fix the c elements,Since \space \alpha\space and \space \beta\space both\space fix\space the\space c \space elements,

(αβ)(ci)=α(β(ci))=α(ci)=ci and(\alpha\beta)(c_{i})=\alpha(\beta(c_{i}))=\alpha(c_{i})=c_{i}\space and

(βα)(ci)=β(α(ci))=β(ci)=ci.(\beta\alpha)(c_{i})=\beta(\alpha(c_{i}))=\beta(c_{i})=c_{i}.

thus αβ(x)=βα(x) x  Sthus \space \alpha\beta(x)=\beta\alpha(x)\forall\space x\space \in\space S


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