An example of an abelian group which is not cyclic

G=Z6×Z2G=Z6×Z2 will do (where ZnZn denotes the cyclic group of order nn). As a direct product of cyclic (so abelian) groups, GG is again abelian. Given any element (x,y)∈G(x,y)∈G, the order of (x,y)(x,y) will be the least common multiple of the orders of x,y.x,y.  

The order of xx must divide 66 and the order of yy must divide 2,2, so the order of (x,y)(x,y) is at most lcm(6,2)=6.lcm⁡(6,2)=6. But |G|=|Z6|⋅|Z2|=6⋅2=12.|G|=|Z6|⋅|Z2|=6⋅2=12. Since no element of GG has order 12,12, then GG is not cyclic.

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )

answer:   2 multiplied by x =24

step-by-step explanation: