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Let G be the cyclic group generated by x, say. Ten every elt of G is of the form x^a, for some a

Q: Prove that every a cyclic group is an abelia?

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There's a theorem to the effect that every group of prime order is cyclic. Since 5 is prime, the assertion in the question follows from the said theorem.

Yes. The only group of order 1 is the trivial group containing only the identity element. All groups of orders 2 or 3 are cyclic since 2 and 3 are both prime numbers. Therefore, any group of order less than or equal to four must be a cyclic group.

By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.

Four of them.

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Yes, every subgroup of a cyclic group is cyclic because every subgroup is a group.

every abelian group is not cyclic. e.g, set of (Q,+) it is an abelian group but not cyclic.

There's a theorem to the effect that every group of prime order is cyclic. Since 5 is prime, the assertion in the question follows from the said theorem.

No.

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

No! Take the quaternion group Q_8.

A cyclic group of order two looks like this.It has two elements e and x such that ex = xe = x and e2 = x2 = e.So it is clear how it relates to the identity.In a cyclic group of order 2, every element is its own inverse.

Yes. The only group of order 1 is the trivial group containing only the identity element. All groups of orders 2 or 3 are cyclic since 2 and 3 are both prime numbers. Therefore, any group of order less than or equal to four must be a cyclic group.

A cyclic group, by definition, has only one generator. An example of an infinite cyclic group is the integers with addition. This group is generated by 1.

No Q is not cyclic under addition.

By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.

The order of a cyclic group is the number of distinct elements in the group. It is also the smallest power, k, such that xk = i for all elements x in the group (i is the identity).