Answer to Question #223550 in Abstract Algebra for Prathibha Rose

Question

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

Answer 1

$Suppose \space \alpha \space and\space \beta\space are \space permutations \space of\\ S= a_{1}, a_{ 2}, . . . , a_{m}, b_{1}, b_{2}, . . . , b_{n}, c_{1}, c_{2}, . . . , c_{k}\\where \space the\space c’s \space are \space left \space fixed \space by \space both \space \alpha \space and \space \beta$

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

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

$(\alpha\beta)(a_{i})=\alpha(\beta(a_{i}))=\alpha(a_{i})=a _{i+1}(with\space a_{m+1}=a_1)and$

$(\beta\alpha)(a_{i})=\beta(\alpha(a_{i}))=\beta(a_{i})=a _{i+1},\\so\space \alpha\beta=\beta\alpha \space on\space the\space a \space elements.$

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

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

$(\alpha\beta)(c_{i})=\alpha(\beta(c_{i}))=\alpha(c_{i})=c_{i}\space and$

$(\beta\alpha)(c_{i})=\beta(\alpha(c_{i}))=\beta(c_{i})=c_{i}.$

$thus \space \alpha\beta(x)=\beta\alpha(x)\forall\space x\space \in\space S$