Question

5. 20 points) Let n 3 be an integer. Let S, be the permutation group on n letters. a) (10 points) Find explicitly , TE S, such that T *o-To. Prove your answer. c) (10 points) Find explicitly an injective map f: (Z3, +a) →S, which satisfies the homomorphism property. Prove your answer.

Answer 1

take o. (12) TE (23) ot= (12) EYESALANAN then (12) (23) (12) = c olto R (13) #T BENEV fi (ziz, tz) — Son f(1) = (123) f(2)= (132) f(0) = id where id is identity. Since if xey & f(x) + fly) for given map. so. f is injective To Prone homomorphism. How f 1 a+b) = Fla) #flb) f (0) = id To Show flo+1) • FIO) * FU) - flote) = flo) * F(*) — 1 f(070) - Flo) * Flo) – Fli+1) = f( * flu) - 9 Fle) = f(1+2) = f(1) * F(2) - where * is oberthion in group Sn. Flito) = f(1) #flo) –

Riau FL2+0) - Fl2)* Flo) - F(2+1) = f(2) * Fli)-9 f(1) = f(2+2) = f(d) a Fle) - 0 & Start from first L.H.S flo+1) = f1) = (123) (L.His means Left Hand side) - RoHS " Rils fo) * FLI) = id. (123) - (123) So kihisa Rilis for Now varity # LIHS of ☺ is f(0+2) = f(2)= (132) R.H.S of ☺ is flo) a f(2)= id. (13.2)= (132) variting & L'Hš of 3 is floto) = flola id Rott's of 3 i Flo) *fla)= id.id = id. vasining | of ☺ is f(1+1) = f fle Ha) = (132) L.H.S of 2. HS of u in F(*) * Fl1)- (123) (123)= (132) variting © LHS of is fll+8)= flo)= id Roths of ☺ is f (1) *F (): 1133) (132) = id variting ③ Luhrs of 0 Flit0) = Flu) = (123) RoH.S. of ( is F (1) * FloD: (193).id = (123)

L.H.S of is Fla+o)- FI29 = (139) Ritts of ☺ us F(2) * Flo) (13.9).id = (132) L'HiS of in F1 2+1) - Flo) id RoHS of ③ is F (9)«Fl)- (132) (123)= id L.HiS of 6 s +(2+2) = f(!) = (123) RoHS of 6 is f(29.61295 (132) (132) = (123) Hence in each case fluty) = f(x) a fly) f(o)= id Hence of is a group homomorphism.

Answer 2

57 We are considering Sor (273). (a) To find o Ce Sn, so that e foter Let C = (12) and r = (1 2 3 4 -- 0-1), N) i.c., In t, 1 goes to 2 2 3 3 & 4 4 - - - - - -D " " n nu 1 n " (81) in " (n-1) " (01-29 4 u 13 3 m 2 2 1 (m-1 .-- . 4 3 2 je, o = (1 n

Date. = 1 n (0-1) -- -- 4 3 2) (1 2) (1 2 3. n) (1n con-lg --- 4 32) (2 3 4 ... 6-1) n) 2 (12) tn 7,3 Exp! Let Sz (n=3 here) E = (12) T=(123) Then th=(132) So toler (132) (1 2) (123) .(1 322 (23) Hence for In T= C = (12) 4 .- - (n-1) n C1 2 3

Date. (c) To find an injective map of : (743, +3) > Su , satisfying (n3) homoror bhi.com property Th3 = ? 5 T , E? f(o) se (identity permutation or trivial permutation) f(T) = (123) f () - CI 3.29 Clearly of is injective mapping, f ODPD) - C123) (13.23 ( 2 3 1 3 1 2 ORD | Hence of (T + 2) = f(T) f (2)

Date. to + I) F() > CL 23) fofo se.(123) C 2 3 U 2 12 31 2 3 1 3 2 3 1 = (1 2 3) sy flő +1) = f (5) f D) Similarly f ( 5 + 2) = f(2)=(132) foof (D) se.(1 32)o(1 329 of f (5+2) = f(0) of (2) Hence ta, bt Zz, +3), f(a.b) = f(a) f(b) je of satisfies' homomorphison property.