7 = the unit group of the ring **Z7 is a cyclic group** with generator 3.

Contents

- Is Z 7Z cyclic?
- Is Z7 a group?
- Is Z6 a cyclic group?
- Is Z * ZA cyclic group?
- Is Z 2Z cyclic?
- Is Z6 abelian?
- What is Z7 in algebra?
- What does Z7 mean in maths?
- Is Z7 a group under multiplication modulo 7?
- Is z5 a cyclic group?
- Is Z8 a cyclic group?
- What are the generators of Z7?
- What is Z6 group?
- Is Z3 cyclic?
- Is Z2 Z6 cyclic?
- Is Z2 Z3 cyclic?
- Is Z nZ * cyclic?
- Is Z 2Z a ring?
- What is z7 isomorphic to?

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

(c) Z7 under multiplication. This is **not a group** because element 0 has no inverse.

Now we see that Z6 = 〈 1 〉, so **Z6 is cyclic** and since every subgroup of a cyclic group is cyclic, we have found all the subgroups (there are no non-cyclic subgroups, we so didn’t miss any). Thus the (distinguishable) subgroups of Z6 are 〈 0 〉, 〈 3 〉, 〈 2 〉 and Z6.

The set of integers Z forms a group with the addition operation. **It is an infinite cyclic group** because all integers can be written by repeatedly adding or subtracting the single number 1.

So (Z/2Z) × (Z/2Z) is **not cyclic**. There is the following simple criterion for when a finite group is cyclic: Lemma 2.7. Let G be a finite group with #(G) = n.

On the other hand, **Z6 is abelian** (all cyclic groups are abelian). So S3 ∼ = Z6.

Z7 = **{0,1,2,3,4,5,6}** Addition (and then multiplication) are defined as in ordinary arithmetic. except that we remove multiples of 7 until we get back to set Z7. So 3+6=8=8 − 7=1. and 3.6 = 18 = 18 − 2.7 = 4 and 34 = 81 = 81 − 11.7 = 4.

Z7 usually denotes **the set of integers modulo 7**. This means that you break up the integers into equivalence classes. You do this by looking at each integer and checking what the remainder is when you divide by 7.

4. [1 points] **Z7 = 1 2 3 4 5 6 is the group of invertible congruence classes modulo 7 under multiplication**.

**It is the fifth order cyclic group**. It is the additive group of the array of five elements.

**Z8 is cyclic of order 8**, Z4 × Z2 has an element of order 4 but is not cyclic, and Z2 × Z2 × Z2 has only elements of order 2. It follows that these Groups are distinguishable. In fact, there are 5 different groups of order 8; the remaining two are non-Abelian.

The “same” group can be written in multiplicative notation like this: Z7 = {1, a, a2, a3, a4, a5, a6}. In this form, **a is a generator of Z7**. It turns out that in Z7 = {0, 1, 2, 3, 4, 5, 6} any non-zero element generates the group. On the other hand, in Z6 = {0, 1, 2, 3, 4, 5} only 1 and 5 are produced.

**One of the two groups of order 6, which is in contrast to abelian**. It’s also a cyclical. It is isomorphic to . Examples are the point groups and , the integers modulo 6 under addition, and the modulo multiplication groups , , and .

(d) • **Z3 is cyclic**, generated additively by 1: we have [1], and [1] + [1] = [1+1] = [2] and [1 ]+[1]+[1] = [1+1+1] = [0] = e, so all elements are captured. is cyclic, multiplicatively generated by [5]: we have [5], and [5]2 = [1] = e. Z2 × Z2 is not cyclic: there is no generator.

This might help to make your problem a bit clearer, noting that **Z2, Z3 and Z6≅Z2×Z3 are each cyclic**, but Z2×Z2 of order 4 is not cyclic. In fact, there is one and only one group of order 4 isomorphic to Z2×Z2, i.e. H. the small 4 group.

It follows that **Z2 × Z3 is a cyclic group with generator (1,1)** such that Z2 × Z3 is isomorphic to Z6. Zmi 2 Page 3 is cyclic if and only if they are pairwise prime. α(1,1) = (α, α), for α any integer.

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

Consider the two rings Z and 2Z. These are **isomorphic as groups** because the function Z −→ 2Z sending n −→ 2n is a group homomorphism that is one to one and on. However, φ is not an isomorphism of rings (in fact, they are not isomorphic like rings).

Group Theory – Show that Z(7)\0 is isomorphic to **C(6)** under modular multiplication – Mathematics Stack Exchange.

- https://www.math.upenn.edu/~chai/370f09/370hwf09/Math370-3.pdf
- http://math.bu.edu/people/rpollack/Teach/541fall09/HW1_Solutions.pdf
- https://billcookmath.com/exams/math3110/math3110-spring2009-test2-answer_key.pdf
- https://en.wikipedia.org/wiki/Cyclic_group
- http://www.math.columbia.edu/~rf/subgroups.pdf
- https://web.northeastern.edu/suciu/MATH3175/MATH3175-fa10-sol4.pdf
- http://www.maths.usyd.edu.au/u/UG/SM/MATH3062/r/lect1.pdf
- https://math.stackexchange.com/questions/2604977/what-does-bbb-z-7-mean
- https://www.math.lsu.edu/~adkins/m4200/4200f09ex2a.pdf
- https://groupprops.subwiki.org/wiki/Cyclic_group:Z5
- https://sites.millersville.edu/bikenaga/abstract-algebra-1/product/product.pdf
- https://sites.millersville.edu/bikenaga/abstract-algebra-1/cyclic-groups/cyclic-groups.pdf
- https://archive.lib.msu.edu/crcmath/math/math/f/f157.htm
- https://www.maths.tcd.ie/~zaitsev/1214-2013/ex1214-7-sol.pdf
- https://math.stackexchange.com/questions/340067/how-to-find-subgroups-of-bbb-z-2-times-bbb-z-6
- http://www.math.ucsd.edu/~jmckerna/Teaching/15-16/Spring/103B/l_1.pdf
- http://ramanujan.math.trinity.edu/rdaileda/teach/s18/m3341/ZnZ.pdf
- http://www.math.ucsd.edu/~jmckerna/Teaching/15-16/Spring/103B/l_10.pdf
- https://math.stackexchange.com/questions/2006904/show-that-z7-0-under-modular-multiplication-is-isomorphic-to-c6

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