APP. B] ALGEBRAIC SYSTEMS 453
(c) We have 2^2 =4, 2^3 =8, 2^4 =1. Hence| 2 |=4 and gp(2)={1, 2, 4, 8}. Also,
72 =4, 7^3 = 4 ∗ 7 =13, 7^4 = 13 ∗ 7 =1. Hence| 7 |=4 and gp(7)={1, 4, 7, 13}. Lastly, 11^2 =1. Hence| 11 |= 2
and gp(11)={1, 11}.
(d) No, since no element generatesG.B.8.Consider the symmetric groupS 3 whose multiplication table is given in Fig. B-4.(a) Find the order and the group generated by each element ofS 3.
(b) Find the number and all subgroups ofS 3.
(c) LetA={σ 1 ,σ 2 }andB={φ 1 ,φ 2 }. FindAB,σ 3 A, andAσ 3.
(d) LetH=gp(σ 1 )andK=gp(σ 2 ). Show thatHKis not a subgroup ofS 3.
(e)IsS 3 cyclic?(a) There are six elements: (1)ε, (2)σ 1 , (3)σ 2 ,(4)σ 3 , (5)φ 1 , (6)φ 2. Find the powers of each elementxuntilxn=ε.
Then|x|=n, andgp(x)={ε,x^1 ,x^2 ,...,xn−^1 }. Notex^1 =x, so we need only begin withn=2 whenx=ε.
(1) ε^1 =ε;so|ε|=1 andg(ε)={ε}.(2) σ 12 =ε; hence∣∣
σ 1∣∣
=2 andgp(σ 1 )={ε, σ 1 }.(3) σ 22 =ε; hence∣
∣σ 2
∣
∣=2 andgp(σ 2 )={ε, σ 2 }.(4) σ 32 =ε; hence∣∣
σ 3∣∣
=2 andgp(σ 3 )={ε, σ 3 }.(5) φ 12 =φ 2 ,φ^31 =φ 2 φ 1 =ε; hence∣
∣φ 1
∣
∣=3 andgp(φ 1 )={ε, φ 1 ,φ 2 }.(6) φ 22 =φ 1 ,φ^32 =φ 1 φ 2 =ε; hence∣∣
φ 2∣∣
=3 andgp(φ 1 )={ε, φ 2 ,φ 1 }.(b) First of all,H 1 ={ε} andH 2 =S 3 are subgroups ofS 3. Any other subgroup ofS 3 must have order 2 or 3 since
its order must divide∣
∣S 3
∣
∣=6. Since 2 and 3 are prime numbers, these subgroups must be cyclic (Problem B.61)
and hence must appear in part (a). Thus the other subgroups ofS 3 follow:H 3 ={ε, σ 1 },H 4 ={ε, σ 2 },H 5 ={ε, σ 3 },H 6 ={ε, φ 1 ,φ 2 }Accordingly,S 3 has six subgroups.
(c) Multiply each element ofAby each element ofB:σ 1 φ 1 =σ 2 ,σ 1 φ 2 =σ 3 ,σ 3 φ 1 =σ 3 ,σ 2 φ 2 =σ 1Hence AB={σ 1 ,σ 2 ,σ 3 }.
Multiplyσ 3 by each element ofA:σ 3 σ 1 =φ 1 ,σ 3 σ 2 =φ 2 , hence σ 3 A={φ 1 ,φ 2 }Multiply each element ofAbyσ 3 :σ 1 σ 3 =φ 2 ,σ 2 σ 3 =φ 1 , hence Aσ 3 ={φ 1 ,φ 2 }(d) H={e,σ 1 },K={e,σ 2 } and thenHK={e, σ 1 ,σ 2 ,φ 1 }, which is not a subgroup ofS 3 sinceHKhas four
elements.
(e) S 3 is not cyclic sinceS 3 is not generated by any of its elements.