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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.
What is finite and infinite cyclic group?
Normally, a cyclic group is defined as a set of numbers generated by repeated use of an operator on a single element which is called the generator and is denoted by g.If the operation is multiplicative then the elements are g0, g1, g2, ...Such a group may be finite or infinite. If for some integer k, gk = g0 then the cyclic group is finite, of order k. If there is no such k, then it is infinite - and is isomorphic to Z(integers) with the operation being addition.
Is a group of order 24 abelian group or not?
The abelian groups of order 24 are C3xC8, C2xC12, C2xC2xC6. There are other 12 non-abelian groups of order 24
Example of group is an abelian group?
The set of integers, under addition.
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.
Is every abelian group is cyclic or not and why?
every abelian group is not cyclic. e.g, set of (Q,+) it is an abelian group but not cyclic.
Is every abelian group is cyclic or not?
No.
Prove that a group of order three is abelian?
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.
Can a non-abelian group have a torsion subgroup?
Yes, a non-abelian group can have a torsion subgroup. A torsion subgroup is defined as the set of elements in a group that have finite order. Many non-abelian groups, such as the symmetric group ( S_3 ), contain elements of finite order, thus forming a torsion subgroup. Therefore, the existence of a torsion subgroup is not restricted to abelian groups.
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.
Every subgroup of a cyclic group is cyclic?
Yes, every subgroup of a cyclic group is cyclic because every subgroup is a group.
Is every solvable group abelian?
No.
If PS are distinct primes show that an abelian group of order PS must be cyclic?
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.
What is finite and infinite cyclic group?
Normally, a cyclic group is defined as a set of numbers generated by repeated use of an operator on a single element which is called the generator and is denoted by g.If the operation is multiplicative then the elements are g0, g1, g2, ...Such a group may be finite or infinite. If for some integer k, gk = g0 then the cyclic group is finite, of order k. If there is no such k, then it is infinite - and is isomorphic to Z(integers) with the operation being addition.
Is the symmetry group of the square an abelian group?
Abelian meaning commutative. If the symmetry group of a square is commutative then it's an abelian group or else it's not.
What is the definition of an abelian group?
An abelian group is a group in which ab = ba for all members a and b of the group.
Prove that every a cyclic group is an abelia?
Let G be the cyclic group generated by x, say. Ten every elt of G is of the form x^a, for some a
