every abelian group is not cyclic. e.g, set of (Q,+) it is an abelian group but not cyclic.
No.
Yes, every subgroup of a cyclic group is cyclic because every subgroup is a group.
No.
Abelian meaning commutative. If the symmetry group of a square is commutative then it's an abelian group or else it's not.
If ( G ) is an abelian group of order ( PS ), where ( P ) and ( S ) are distinct primes, then by the Fundamental Theorem of Finite Abelian Groups, ( G ) can be expressed as a direct product of cyclic groups of prime power order. The possible structures for ( G ) are ( \mathbb{Z}/PS\mathbb{Z} ) or ( \mathbb{Z}/P^k\mathbb{Z} \times \mathbb{Z}/S^m\mathbb{Z} ) with ( k ) and ( m ) both ( 1 ). However, since ( P ) and ( S ) are distinct primes, the only way for ( G ) to maintain order ( PS ) while being abelian is for it to be isomorphic to ( \mathbb{Z}/PS\mathbb{Z} ), which is cyclic. Thus, ( G ) must be cyclic.
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
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.
Yes, every subgroup of a cyclic group is cyclic because every subgroup is a group.
No.
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.
Abelian meaning commutative. If the symmetry group of a square is commutative then it's an abelian group or else it's not.
An abelian group is a group in which ab = ba for all members a and b of the group.
If ( G ) is an abelian group of order ( PS ), where ( P ) and ( S ) are distinct primes, then by the Fundamental Theorem of Finite Abelian Groups, ( G ) can be expressed as a direct product of cyclic groups of prime power order. The possible structures for ( G ) are ( \mathbb{Z}/PS\mathbb{Z} ) or ( \mathbb{Z}/P^k\mathbb{Z} \times \mathbb{Z}/S^m\mathbb{Z} ) with ( k ) and ( m ) both ( 1 ). However, since ( P ) and ( S ) are distinct primes, the only way for ( G ) to maintain order ( PS ) while being abelian is for it to be isomorphic to ( \mathbb{Z}/PS\mathbb{Z} ), which is cyclic. Thus, ( G ) must be cyclic.
Let G be the cyclic group generated by x, say. Ten every elt of G is of the form x^a, for some a
The abelian groups of order 24 are C3xC8, C2xC12, C2xC2xC6. There are other 12 non-abelian groups of order 24
An abelianization is a homomorphism which transforms a group into an abelian group.