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Thus, D4/Z (D4) is an abelian group of order 4. Now, by the fundamental theorem of finite abelain groups, D4/Z (D4) is either isomorphic to Z4 or Z2 X Z2. But Z4 is cyclic, and if D4/Z (D4) was isomorphic to Z4, then this would imply that D4/Z (D4) is also cyclic, which would imply D4 is abelian (by elementary theorem). This is clearly not true.
Z6 5 generates this group: 52=7, 53=17,54=13,55=11, and 56=1. 20 {1,3,7,9,11, 13,17,19} (Z2)x(Z4) Here is an isomorphism. f(0,1)=3. and 34=1. So, f(0,2)=9, f(0,3)=7 and f(0,0)=1. 11*3=13. Let f(1,1)=13. 11*9=19. Let f(1,2)=19. 11*7=17. Let f(1,2)=17. 21 {1,2,4,5,8,10,11, 13,16,17,19,20} (Z2)x(Z6) Let f(1,0)=13; note that 132=1.
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