Sage-Tips

How do you prove a cyclic is a finite group?

Let G be a group. If the order of G is a prime, or of the form pq where p numbers such that pdoes not divide , then G is cyclic.

How do you know if a subgroup is cyclic?

Theorem: All subgroups of a cyclic group are cyclic. If G=⟨a⟩ is cyclic, then for every divisor d of |G| there exists exactly one subgroup of order d which may be generated by a|G|/d a | G | / d . Proof: Let |G|=dn | G | = d n .

How do you identify a cyclic group?

Starts here9:00(Abstract Algebra 1) Definition of a Cyclic Group – YouTubeYouTube

Are all cyclic groups finite?

Every cyclic group is virtually cyclic, as is every finite group. An infinite group is virtually cyclic if and only if it is finitely generated and has exactly two ends; an example of such a group is the direct product of Z/nZ and Z, in which the factor Z has finite index n.

Does an infinite group have an order?

There are infinitely many rational numbers in [0,1), and hence the order of the group Q/Z is infinite. Thus the order of the element mn+Z is at most n. Hence the order of each element of Q/Z is finite. Therefore, Q/Z is an infinite group whose elements have finite orders.

What is the order of a finite group?

The number of elements of a group (finite or infinite) is called its order. We denote the order of G by |G|. Definition (Order of an Element). The order of an element g in a group G is the smallest positive integer n such that gn = e (ng = 0 in additive notation).

How to prove that H is unique in a finite cyclic group?

Let G = ⟨ x ⟩ be a finite cyclic group of order n and let d be a divisor of n. Then x n d has order d and so it generates a subgroup H of G of order d. To establish that H is unique, assume that K is any subgroup of order d in G. Since G = ⟨ x ⟩, K = ⟨ x m ⟩ for some integer m, and so x m has order d.

What is the unique subgroup of G of order 1?

Let G 1 be the subgroup of order p m − 1. Now, p-groups always have at least one subgroup of each possible order. (This general theorem follows from the fact that P contains a central subgroup of order p, plus induction.) Therefore G 1 contains the unique subgroup of G of order 1, p, p 2, …, p m − 1.

How many different subgroups are there in a cyclic group?

To help you understand where you’re going wrong, why not try writing out these “six different subgroups”: if $G$ is a cyclic group of order $7$, and $a$ is a generator of $G$, then

Is every subgroup of a ∗ -group normal?

Note also that in a group satisfying ∗, every subgroup is the unique subgroup of its order and in particular, is equal to each of its conjugates. In other words, every subgroup of a ∗ -group is normal. We first prove that any finite cyclic group has property ∗.