Question

4. (a) (3 points) List all the subgroups of the symmetric group S3. (b) (4 points) List all the normal subgroups of Sz. (c) (3 points) Show that the quotient of S3 by any nontrivial normal subgroup is a cyclic group.

Answer 1

theowm, have Q@ Sz={ (1) (2) (3), (1 2)(3), (13)(2), The order of S3 = 31=6, so by Lagrange's (2 3)(4), (1 2 3 ), ( 3 2) Order TH= 1, 2, 3,6. The extremel a subgroup Hq Sz can only correspond to the trivial subgroups H=(e) and H= 53. Case I. 141=3. The alternating group A3 seg us two 3-cyclesy is a normal Subgroup of order 6/2=3. Obviously A3 = Z3 since there is only group of order 3. For if TH1=3 then HAA3 is a subgroup in both Az and H and by Lagrange's theorem can only have order 3 (in which case. H= or 1 (and then Azn H=(e)}. The later possibility Can not arise. If it did, then Counting principle the product set would have cardinality. I HA3] =#1. 1 Asl fermos =g whink exceeds '1 $3] = 6. Only Case runains. Case 2. 141=2. The cycle The cycle types in Sy are cycle Type Example #(Elements in S3 1 any 2-cycle any 3-cycle 3 She e 11 12 3 3 2

Thus all are (1)(2)(3), (123), (132)ý = A3 {(1), (2) (3)y=e subgroups of S3 e he have ghg I E Azi nigt. This weonly ş (1) (2)(3), (1.2) (3) $(1)(2)(3), (1 3) (2) {(1)(2)(3), (2 3) (1) 9. (b). We observed that the unique subgroup of Order 3 in S3 is Az . It remains to show that for any ge € Sz and he A3 We make the observation that (ghgt) (ghgt) gh²gt. need to check g(123)gte te A3 forall g E. Sz; since (132) (123)2 and A3 is closed under the group operation, if g (123) g7 Edz, EA3, then g (132) gt (123) We do the necessa cheeks calculations (1 2) (123) (12) = (1 32) EAz (1 3) (123) (13) = (132) € Az (2 3] (123)(23) = (1 32) Az We have shown that Az is a normal subgroup of S3. We can check that other three -L g (123)?g Az gtje by direct

are n not normal Conside for (103) € S3 we have are Similarly re ting non trivial subgroups of S3 (1) (2) (3).(1 2)(3) (13) (12) (1 3) = (2 37€ HI remainin two subgroups not namel (c) Here S3 / A3 contains exactly elements and S3 / A3 E Zz This S/ Az has two elements, first elements is identity and other element is generator of subgroup. Therefore {/ A3 cjclic two يا