Yes, every subgroup of a cyclic group is cyclic because every subgroup is a 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.
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 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.
every abelian group is not cyclic. e.g, set of (Q,+) it is an abelian group but not cyclic.
Four of them.
There are two: the cyclic group (C10) and the dihedral group (D10).
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.
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.
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.
No.
No Q is not cyclic under addition.