0
What else can I help you with?
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 group whose order is less than or equal to 4 a cyclic group?
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.
Prove that a group of order 5 must be 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.
A cyclic group of length 2 is called identity?
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.
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.
Let G be a cyclic group of order 8 then how many of the elements of G are generators of this group?
Four of them.
How do you determine number of isomorphic groups of order 10?
There are two: the cyclic group (C10) and the dihedral group (D10).
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.
Is it true that an infinite cyclic group may have 3 distinct generators?
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.
Show that an element of a group has order n if and only if it generates a cyclic group of order n?
An element ( g ) of a group ( G ) has order ( n ) if the smallest positive integer ( k ) such that ( g^k = e ) (the identity element) is ( n ). This means the powers of ( g ) generate the set ( { e, g, g^2, \ldots, g^{n-1} } ), which contains ( n ) distinct elements. Therefore, the cyclic group generated by ( g ), denoted ( \langle g \rangle ), has exactly ( n ) elements, thus it is a cyclic group of order ( n ). Conversely, if ( \langle g \rangle ) is a cyclic group of order ( n ), then ( g ) must also have order ( n ) since ( g^n = e ) is the first occurrence of the identity.
How many groups have only 5 elements in them?
In group theory, there is exactly one group of order 5, which is the cyclic group ( \mathbb{Z}/5\mathbb{Z} ). This is because 5 is a prime number, and any group of prime order is cyclic and isomorphic to the integers modulo that prime. Therefore, up to isomorphism, there is only one group with 5 elements.
How do we prove that a finite group G of order p prime is cyclic using Lagrange?
Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.
