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

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Q: Every subgroup of a cyclic group is cyclic?
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Proof or Disprove 'If every proper subgroup of G is cyclic then G must be cyclic'?

No! Take the quaternion group Q_8.


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.


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.


Is every abelian group is cyclic or not?

No.


In group theory what is a group generator?

In abstract algebra, a generating set of a group Gis a subset S such that every element of G can be expressed as the product of finitely many elements of S and their inverses.More generally, if S is a subset of a group G, then , the subgroup generated by S, is the smallest subgroup of G containing every element of S, meaning the intersection over all subgroups containing the elements of S; equivalently, is the subgroup of all elements of G that can be expressed as the finite product of elements in S and their inverses.If G = , then we say S generatesG; and the elements in S are called generators or group generators. If S is the empty set, then is the trivial group {e}, since we consider the empty product to be the identity.When there is only a single element x in S, is usually written as . In this case, is the cyclic subgroup of the powers of x, a cyclic group, and we say this group is generated by x. Equivalent to saying an element x generates a group is saying that it has order |G|, or that equals the entire group G.My source is linked below.