Question

please provide the correct answer to #2, please write clearly, thank you

please support the abswer with appropriate reasoning.

Oooh OOCOUPO 2. Please list all factor groups and subgroups of the cyclic group C6 of order 6. Is there any other abelian group of this order which is not isomorphic to C6?

Answer 1

do a PP "Ist we find subgroub of Con{0, 1, 2, 3, 4, 5%. Since under operation muliplication mod 6. 6 is Cyclic so each subgroup of c are Cyclic [as subgroup of cyclic group is cycle]. Also for each divisor of 6 , It has a subgroup of that order and that is subgroup in unique (as each Eyclic group of finite order (say n is unique upto iromorphilm namely so divisor of 6 are 1, 2, 3, 6. so o has subgroups are of orderte 1,2,3 and 6 genereted by element 1,2,3, and 6 respectively. So there are y subgroups only (say Hi genereted by it i=1,2,3,6). Hi= { not (mod 6)/n. }. (generated by 4). = {0, 1, 2, 3, 4, 5, 6} (co itself). H2= {v.2 (mod6) / ne 28. (generated by 2). = {0, 2, 4 H2= { v. 3 (mod 6) Int2} (genereted by 3). = {0,3}. Ho = {nob (mod 6) /nt4} (genereted by 6). u 10} Here out of these y subgroup of Co No subgroup of G is lett, as we knows the subgroups are are of finire order (as o(66) = 6(finite) and order of subgroup divides order of group and upto isomorphism each subgroup is a

UP A P P P P unique (by ernation ) Now Here we discuss factor group of Co. We know that, the factor group of Glang Group are of the form. W = {athlath}. where His Normal Subgroup of G. Also, we know. Every subgroup of Abelian group is Normal Subsob ll are of that group! the Bubgroup normal 2, C6 il Cycle Hence Coin Abelian). so All factor group of Co are of the form. 27 | 22 VAUDAMA = {a+HilaEco} = {0+ H, 4+ Hp, 24H2, 3+421 4+H, st Higa He has at H = Hift.( ) atH). S = {0+ H₂, It H₂, 3 + H₂, 5th2} (as 2+H₂ = of H₂ 1 H2 Y + H= 0+Hz. We knew . similirly atha bt Ha 6 = {0th 3, lt Hz, 24H 3, Y+H 3, 5+ Hy} arbeta . I --- So there are only y Normal suby roup.

We know that up to isomorbhim, there are only two group of order 6.0 one is co and other is C₂ X Cz Cz xq is Abelian group of order 6 which is not a inomorphic to 76. and C₂ X q = {(0,0), (0, 1), (012), (2 رار(۱) زهرا) (20), (211), (2, 2)}. under operation (a,b) (c,d) = (atc (mod 6), elle (mos Cox G is not isomorphic to co as CzX Cz does not have element of order 6 by C has element of onder 6 015)= 0(1)26. o they are not ixomorphic.