Is Z nZ a cyclic group?

(Z/nZ,+) is cyclic since it is generated by 1 + nZ, i.e. a + nZ = a(1 + nZ) for any a ∈ Z.

Are all Z groups cyclic?

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.

Which are the invertible elements of Z?

Modular Arithmetic Recall: Let n ∈ Z such that n > 0 and a ∈ Zn. An element a ∈ Zn = {0,1,…,n − 1} is invertible if and only if gcd(a, n) = 1. 1 = a · x + n · y • If 1 = a · x + n · y and b = x mod n, then a ⊗ b = 1. 1.

Is Z 16Z cyclic?

There is no such cyclic group, since (Z/16Z)× is isomorphic to Klein’s Vierergruppe, i.e. it is the product of a cyclic group of order 2 and a cyclic group of order 4, since its elements are: ±1,±3,±5=∓33,±7=∓32.

Why is Z 8Z not cyclic?

But (Z/8Z)∗ is not cyclic, as 32 = 52 = 72 = 1. Hence every element of (Z/8Z)∗ has order 1 or 2. In particular, there is no element of (Z/8Z)∗ of order 4, so that (Z/8Z)∗ is not cyclic. Thus φ(n) is the number of a ∈ Z, 0 ≤ a ≤ n − 1, such that gcd(a, n) = 1.

Is z * z cyclic?

So we have n = 0. Consider the element (n,−m) ∈ Z × Z. There is an integer k ∈ Z with (kn, km)=(n,−m), and since n, m = 0 this gives k = 1 and k = −1, which is a contradiction. So Z × Z cannot be cyclic.

How do I know if a group is cyclic?

A finite group is cyclic if, and only if, it has precisely one subgroup of each divisor of its order. So if you find two subgroups of the same order, then the group is not cyclic, and that can help sometimes.

What is the elements of Z?

Zirconium – Element information, properties and uses | Periodic Table.

Is Z 7Z a cyclic group?

(1a) Is (Z/7Z)∗ a cyclic group? ANSWER: Yes, it is generated by (3 + 7Z) for example.

Is Z ∗ 20 a cyclic group?

With similar computations, you can see that none of the elements of Z∗20 generates the group. Hence, it’s not cyclic.