Question

(4) Let X be a nonempty set, and let o E Perm(X). The set of fixed points of o is fix(0) = {x € X : 0(x) = x}. The support of o is supp(o) = {x EX :0() #x}. (a) Prove that fix(o) U supp(o) = X and fix(on supp(o) = 0. (b) Prove that fix(oº) = fix(0) and supp(02) = supp(o). (c) Permutations o and T in Perm(X) are disjoint if, for all x E X, we have o(x) + x implies T(x) = x, and T(x) + x implies o (2) = x. Prove that o and T are disjoint permutations if supp(o) C fix(T) and supp(T) C fix(o). (d) Prove that disjoint pemutations commute.

Answer 1

- Now assume x Solution:- Let X be a non empty set and let se Permax where Perm (x) is the set of all permutations on from maps the set x. ie, set of all one-one onto m X to X. Also fix cod= {xex: 6(x)=xy and Suppconfe: 6tel (a) To prove that fix co) U suppco-)=X. By the definition of fixco) and Suppcos we have if * e fzcc)0 S-Peº theo xe. :. fao) U suppco) SX ne X and take of Perm CX) Then we have either o(x)=x or o(a)ta. either ie, Efixco) or xe suppo le = efexo-NUSRP (-) Thus xa fixco) U suppo) 2 from ☺ fixco) U supp(o)= X. Now prove that fixe con suppco)=$ Assume the contrary that there is an element x Efixcoo suppco) ie, a Efixco) and x € Suppco 6 (x)= x and o(o)tx This is contradiction since G is o(a) must be un unique Thus there is x & fix (6) Supp(o). ce, fixco) o Suppco) = ¢ . have to prove that fixco ²) fix (o) and Supp(oz)= Supplo). to e De 1 a map, a ие. по (6) Here we

Thus Supplo) & Supp(o?). Also This is not true. For example let X={a,b,c,d). Let o(a)= b, OCb)=a, Ce)-c, oldad. Then fix (o)= {c,d). But then 8 (a)=666 (as)=o (6)=a 0°C)=0 (6C6))-0(a)- b. 00) (0) (0-C od)-(old) -o ( dad This a?a=a, o (6)=6, 6(C)= 6, 8 (d)=d. .: fixco²)={a,b,c,d). Here fix (o) & fixco?). We can only say that fixo) e fixlo?) Also of support Let X={a,b,c,d} o(a)=6, 6(b)=0, 0(C)=c, Then Supp (o)= {a,6} . But as above o²(a)=a, 5²6=6,0²c)=(,0(d)-d in the case st (d)zd. .: Supploy=0 We can say that Supp (6²) < Supple) (c) To show that o and I are disjoint when suppco) s fix (?) and Supp(a) s fixco). So Assume Suppco) s fix(2) and Supp(a) s fixco) To prove that are disjoint in the sense of given definition. Let & EX and o(a)ta o and Y Then clearly & E Supp(o). Since Suppco) sfix (a) we have xefixcay

i Hence ♡ (c)= a. Thus 6CV)Ex implies 2C%) = x . Also assume ron) &n. Then clearly x E supp (²) wow Since Supp() sfixco) We have xe fx-). ie, o (O)=x. Thus ?(x) && implies & (x)=x. Thus if suppco) s fixca) and Supply)sfix (6) then for all & EX, we have o (»)#x implies Y(n)=x and 268) # implier o(x)=x. Thus o and Y are disjoint. (d) Finally to prove that if or and I are disjoint permutations then then 62=25 (ie, they commute). let xEX. Now if 6(x)=x and Ycx)=x then o Coe= 0 (200))= 6(x)= x and 76(x)=2660+9)=ToD=X This in this case or=80. Y Cx) + x So we may assume assume either o(a) ex or If () Exc, by the definition of disjoint permutations Zcx)= x and in that case o 2(x) = 6 (26x)=5 2). and 20 (x) = 216(x)) Now 260(R))=0 (2) since if it o not the Case ie, e(o(a) to (*) then rly) #4 where yoo (2) so by definition of disjoi sint permutating we must have oly)=y ie, o 6(x)) = 0 (0)

-H2 and ot? ie, o x==(x) Applying o' on both sides (since os tonto, o enist) :oco (w)=566(x) ie, oo²(x) = 6 (2) ie, o (n) = x (since its = identity This is a contradiction since &(2) is assumed to be not equal to x This {co (n)= o(a) Thus we get or=20. Similar proof follows when Ycm) ## . Thus in all ie, X. cases we have or=%6. o and ? commute.