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Is every finite abelian group is cyclic?
No, for instance the Klein group is finite and abelian but not cyclic. Even more groups can be found having this chariacteristic for instance Z9 x Z9 is abelian but not cyclic
Example of group is an abelian group?
The set of integers, under addition.
Is every cyclic group abelian?
Yes. Lets call the generator of the group z, then every element of the group can be written as zk for some k. Then the product of two elements is: zkzm=zk+m Notice though that then zmzk=zm+k=zk+m=zkzm, so the group is indeed abelian.
If d divides the order of a group does the group has a subgroup of order d?
The general answer is no. Consider A4={(1),(12)(34),(13)(24),(14)(23),(123),(124),(132),(134),(142),(143),(234),(243)}. The subgroups of A4 are: A4, , , , =, =, =, =, {(1),(12)(34),(13)(24),(14)(23)}, {(1)}. The order of A4 is 12, the order of , and is 2, the order of =, =, = and = is 3, the order of {(1),(12)(34),(13)(24),(14)(23)} is 4, and the order of is 1. Clearly there are no subgroups of order 6, but 6 definitely divides the order of A4. The statement is true for all finite abelian groups, and when d is a power of a prime (i.e., when d=pk for a prime p and a non-negative integer k).
What is the order of a group?
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
