hw4 - Math 113 Homework # 4, selected solutions 2: (a) Let...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 113 Homework # 4, selected solutions 2: (a) Let i ∈ { 1 ,...,n } ; we must show that f ( g ( i )) = g ( f ( i )). Case 1: i is not in the support of f or g . Then f ( g ( i )) = f ( i ) = i and likewise g ( f ( i )) = g ( i ) = i . Case 2: i is in the support of f . Then f ( i ) is also in the support of f , since otherwise f would send both i and f ( i ) to f ( i ), and hence would not be injective since i 6 = f ( i ). Since i is not in the support of g we have f ( g ( i )) = f ( i ), and since f ( i ) is not in the support of g we have g ( f ( i )) = f ( i ). Case 3: i is in the support of g . By the same argument as in Case 2 but with f and g switched, we have f ( g ( i )) = g ( f ( i )) = g ( i ). (b) Let p be the element of supp( f ) supp( g ). Then you can check that [ f,g ]( p ) = f ( p ), [ f,g ]( f ( p )) = g ( p ), [ f,g ]( g ( p )) = p , and for any x ∈ { 1 ,...,n }\{ p,f ( p ) ,g ( p ) } , we have [ f,g ]( x ) = x . Also p , f ( p ), and g ( p ) are all distinct (why?), so [ f,g ] is the 3-cycle [ f,g ] = ( p, g ( p ) , f ( p )) . 3: (a) Let i ∈ { 1 ,...,n } ; we need to show that σμσ - 1 ( i ) = ( σ ( x 1 ) σ ( x 2 ) ··· σ ( x k ))( i ) . We consider two cases.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

hw4 - Math 113 Homework # 4, selected solutions 2: (a) Let...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online