Use with the (group-theory) tag. A group is cyclic if it can be generated by a single element, $a$. Every element has the form $a^i$ for some $i\in\mathbb{Z}$, and so these groups are abelian.

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8
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3answers
529 views

Why must a field with a cyclic group of units be finite?

Let $F$ be a field and $F^\times$ be its group of units. If $F^\times$ is cyclic show that $F$ is finite. I'm a bit stuck. I know that I can represent $F^\times = \langle u \rangle$ for some $u ...
14
votes
5answers
2k views

Groups of order $pq$ without using Sylow theorems

If $|G| = pq$, $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems?
6
votes
5answers
3k views

Subgroups of a cyclic group and their order.

Lemma $1.92$ in Rotman's textbook (Advanced Modern Algebra, second edition) states, Let $G = \langle a \rangle$ be a cyclic group. (i) Every subgroup $S$ of $G$ is cyclic. (ii) If ...
3
votes
4answers
283 views

$(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic Group for $n\geq 3$

Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic Group for $n\geq 3$. Hint : Find two subgroups of order $2$. I somehow feel that a cyclic group can not have two distinct ...
4
votes
2answers
862 views

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .

I'm looking for a simple proof that up to isomorphism every group of order 2p (p prime) is either $\mathbb{Z}_{2p}$ or $D_{p}$ (The Dihedral group of order 2p). I should note that by simple I mean ...
5
votes
4answers
140 views

A finite group which has a unique subgroup of order $d$ for each $d\mid n$.

Problem Suppose G is a finite group of order $n$ which has a unique subgroup of order $d$ for each $d\mid n$. Prove that $G$ must be a cyclic group. My idea: I try to prove it by induction. Let ...
4
votes
1answer
645 views

G is group of order pq, pq are primes

Problem. Let $G$ be a group of order $pq$ such that $p$ and $ q$ are prime integers. I am to show that every proper subgroup of $G$ is cyclic. My attempt. What I know: Any element $a$ divides $pq$ ...
2
votes
2answers
1k views

If a cyclic group has an element of infinite order, how many elements of finite order does it have?

If a cyclic group has an element of infinite order, how many elements of finite order does it have? I know that the order of the entire group must be infinite, for an element of the group must have ...
6
votes
5answers
2k views

Homomorphism between cyclic groups

I have some confusion in relation to the following question. Let $\langle x\rangle = G$, $\langle y\rangle = H$ be finite cyclic groups of order $n$ and $m$ respectively. Let $f:G \mapsto H$ ...
10
votes
2answers
295 views

Can we conclude that this group is cyclic? [duplicate]

Let $G$ be a finite group. If, for each positive integer $m$, the number of solutions of the equation $x^m = e$ in $G$, where $e$ is the identity element, is at most $m$, then can we conclude that $G$ ...
13
votes
7answers
2k views

Prove that $\mathbb{R^*}$, the set of all real numbers except $0$, is not a cyclic group

Prove that $\mathbb{R^*}$ is not a cyclic group. (Here $\mathbb{R^*}$ means all the elements of $\mathbb{R}$ except $0$.) I know from the definition of a cyclic group that a group is cyclic if ...
5
votes
3answers
2k views

Quadratic subfield of cyclotomic field

Let $p$ be prime and let $\zeta_p$ be a primitive $p$th root of unity. Consider the quadratic subfield of $\mathbb{Q}(\zeta_p)$. For instance, for $p=5$ we get the quadratic subfield to be ...
4
votes
4answers
5k views

Product of two cyclic groups is cyclic iff their orders are co-prime

Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $\gcd(n,m)=1$. I can't seem ...
6
votes
3answers
253 views

For what $n$ is $U_n$ is cyclic?

When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ to be cyclic? $$U_n=\{a \in\mathbb Z_n \mid \gcd(a,n)=1 \}$$ I searched in internet but did not get clear idea.
6
votes
6answers
5k views

Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.

I've tried but I could not find a noncyclic Abelian group all of whose proper subgroups are cyclic. please help me.
5
votes
6answers
6k views

how to find a generator of a cyclic group

A cyclic group is a group that is generated by a single element. That means that there exists an element g, say, such that every other element of the group can be written as a power of g. This ...
2
votes
1answer
217 views

At most one subgroup of every order dividing $\lvert G\rvert$ implies $G$ cyclic [closed]

Suppose we have a finite group $G$ of finite order $n$. For every $d\mid n$, $G$ has at most one subgroup of order $d$. Show that $G$ is cyclic.
1
vote
1answer
188 views

What are the generators for $\mathbb{Z}_p^*$ with p a safe prime?

lets consider $\mathbb{Z}_p^*$ with $p = 2 \cdot q + 1$ a safe prime ($p$ and $q$ have to be prime). Then $\varphi\left(p\right) = 2 \cdot q$ is the order of $\mathbb{Z}_p^*$, and ...
14
votes
3answers
623 views

Show $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$

So I want to show that $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$. My approach to this problem was to show a double containment, i.e. to show that ...
3
votes
3answers
913 views

Showing that a group of order $21$ (with certain conditions) is cyclic

How can i show that if $o(G)=21$ and if $G$ has only one subgroup of order $3$ and only one subgroup of order $7$, then show that $G$ is cyclic.
1
vote
1answer
74 views

Finding homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{6}$.

Find all homomorphisms from $\mathbb Z_{12}$, the cyclic group of order $12$, to $\mathbb Z_6$. For each homomorphism $f\colon \mathbb Z_{12}\to \mathbb Z_6$, determine the kernel $\ker(f)$ and the ...
8
votes
1answer
402 views

How to find all groups that have exactly 3 subgroups?

How to find all groups that have exactly 3 subgroups? Any group must have identity and itself as subgroups, so we just need to find all the groups that only have one proper subgroup. I think ...
1
vote
0answers
52 views

Divisors of the totient function and congruences

Based on the question Question about the totient function and congruence classes, I would like to ask two new questions: Question 1 Does it hold that if $k\mid\varphi(n)$ then the elements of the ...
4
votes
2answers
251 views

Cyclic subgroups and generators

Sorry for posting a second question on this topic, abstract algebra is taking a bit longer to get my head around. I'm trying to work out if this is cyclic or not, and find all generators or show no ...
3
votes
3answers
182 views

A persisting element in all subgroups.

Let $G$ be a finitely generated abelian group and $a$ be a nontrival element of $G$ contained in all nontrivial subgroups of $G$. Is $G$ necessarily cyclic?
2
votes
2answers
87 views

Properties possessed by $H , G/H$ but not G

i) Does there exist a group $G$ with a normal subgroup $H$ such that $H , G/H$ are abelian but $G$ is not ? ii) Does there exist a group $G$ with a normal subgroup $H$ such that $H , G/H$ are cyclic ...
2
votes
1answer
121 views

Cyclic group generators

My question is: Can you find a cyclic group with n generators? I know that zero (or any other identity element for that matter) is included, so there would be for $Z_n$ at most n-1 generators. ...
1
vote
2answers
113 views

Show that the group is cyclic.

I'm trying to show that the group $U(Z_{54})$ is cyclic. To start, I found the divisors of 54 = {1, 2, 3, 6, 9, 18, 27, 54} Then I started to find the elements using the powers of a. Where ...
1
vote
3answers
55 views

Prove that $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{lcm (n,k)} \rangle$

Let $G$ be a group. Let $a$ be an element. Let $n,k$ be pozitive integers. Let $m$ be least common multiple of $n$ and $k$. Prove $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{m} ...
0
votes
1answer
48 views

How do I prove that there exists a cyclic subgroup of order lcm of orders of cyclic subgrpups of an abelian group?

Before I start, please note that this post is not duplicate Let $G$ be an abelian group. Let $H,K$ be finite cyclic subgroups of $G$ such that $|H|=r,|K|=s$. Then, how do I prove that there exists ...
0
votes
3answers
64 views

Prove that if $k\mid m,$ then $Z_m$ has a subgroup of order $k.$

Prove that if $k\mid m$, then $Z_m$ has a subgroup of order $k.$ Ok, so this doesn't look like too hard of a problem. So do I just show that it is closed under multiplication and inverse? I just ...
0
votes
0answers
47 views

Let $G = \langle x\rangle$ be cyclic of order $n$. Prove that $\langle x^r\rangle\leq \langle x^s\rangle$ iff $r$ is a multiple of $s$ modulo $n$.

Let $G = \langle x\rangle$ be cyclic of order $n$. Prove that $\langle x^r\rangle \leq \langle x^s\rangle$ iff $r$ is a multiple of s modulo $n$. Give the complete subgroup lattice of $G = \langle ...
-1
votes
3answers
769 views

Intersection of cyclic subgroups: $(x^m) \cap (x^n) = (x^{lcm(m,n)})$ [duplicate]

This group theory problem has stumped me. I want to prove that if $G=(x)$ is a finite cyclic group that $(x^n) \cap (x^m) = (x^{\operatorname{lcm}(m,n)})$ for all integers $m$ and $n$, where $(x)$ is ...
-3
votes
2answers
84 views

Prove that the subgroup of the quotient group is cycling and infinitely generated

$$M = \left\{\,\dfrac{m}{13^n}\biggm| m\in \mathbb{Z}, n\in\mathbb{N} \,\right\}, \quad G = M/\mathbb{Z}$$ Prove that any subgroup $H < G$, $H\neq G$ is cyclic and infinitely generated and that ...
11
votes
5answers
4k views

Are cyclic groups always abelian?

If a group $C$ is cyclic, is it also abelian (commutative)? If so, is it possible to give an “easy” explanation of why this is? Thanks in advance!
14
votes
1answer
297 views

Fibonacci Sequence in $\mathbb Z_n$.

Consider a Fibonacci sequence, except in $\mathbb Z_n$ instead of $\mathbb Z$: $$F(1) = F(2) = 1$$ $$F(n+2) = F(n+1) + F(n)$$ It is easy to see that each of these sequences must cycle through some ...
12
votes
2answers
5k views

Order of automorphism group of cyclic group

Let $G$ be a cyclic group of order $m$. What is the order of $\text{Aut}(G)$? I want to know the proof as well (elementary if possible). I would still accept the proof if one answers with $m = p$, a ...
10
votes
4answers
2k views

Finite groups with exactly one maximal subgroup

I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated. Let $G$ be a finite group. Suppose ...
6
votes
2answers
3k views

A subgroup of a cyclic group is cyclic - Understanding Proof

I'm having some trouble understanding the proof of the following theorem A subgroup of a cyclic group is cyclic I will list each step of the proof in my textbook and indicate the places that I'm ...
7
votes
1answer
86 views

$n^2$ divides $\phi(a^n-1)$ whenever $n$ is even and $a>2$

My problem is to show that $n^2$ divides $\phi(a^n-1)$ whenever $n$ is even and $a>2$. I have thought a solution but it is quite long and tedious. I wonder if anyone has a nice and clear ...
7
votes
2answers
512 views

Abelian group admitting a surjective homomorphism onto an infinite cyclic group

I am working on the following problem: Let $G$ an Abelian group and $f: G \to \Bbb Z$ a surjective homomorphism. Prove that $G \cong \ker(f) \times \Bbb Z$ By means of the First Isomorphism ...
4
votes
2answers
260 views

Cyclic Group Presentation

Show that the the group with presentation $$\langle x, y\ \mid\ x^2=y^2x^2y,\ (xy^2)^2=yx^2, \ yx^{-1}y^2=x^7\rangle $$ is cyclic of order 24. This presentation was obtained using the Todd-Coxeter ...
9
votes
2answers
235 views

Generators of a cyclic group

In a paper there is a lemma: Let $G= \langle a,b \rangle$ be a finite cyclic group. Then $G=\langle ab^n \rangle$ for some integer $n$. The proof is omitted because it's "straightforward" but ...
6
votes
3answers
649 views

How to find subgroups of $ \;\;\Bbb Z_2\times \Bbb Z_6$

I am reading a first course in algebra and there is an example saying that "find all the subgroups of $\Bbb{Z}_2\times\Bbb{Z}_6$ and decide which of them are cyclic. I know that ...
6
votes
3answers
769 views

On the Factor group $\Bbb Q/\Bbb Z$ [duplicate]

Possible Duplicate: $\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$? I have the factor group $\Bbb Q/\Bbb Z$, where $\Bbb Q$ is group of rational ...
4
votes
5answers
467 views

Prove any subgroup of a cyclic group is cyclic.

was just wondering if this is a valid proof for the aforementioned question? I am quite confident that it isn't, but not exactly sure why. Maybe I am missing the point of proofs by induction ...
2
votes
1answer
2k views

Find all proper nontrivial subgroups of Z2 x Z2 x Z2 - Fraleigh p. 110 Exercise 11.10

$\newcommand{\lcm}[0]{\mathrm{lcm}}$I tried to fill in the steps but I'm confounded by this solution. Here $i$ is the identity element, not $e$. Because $\lcm(2, 2, 2) = 2$ hence all non-identity ...
1
vote
4answers
455 views

Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46.

This solution is from here and yahoo. Given $a,b$ elements of $G$, and $ab$ has finite order $n$. Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$. Need to show $n$ is the smallest positive integer ...
1
vote
3answers
223 views

Intuition and Tricks - Hard Overcomplex Proof - Order of Subgroup of Cyclic Subgroup - Fraleigh p. 64 Theorem 6.14

Update Dec. 28 2013. See a stronger result and easier proof here. I didn't find it until after I posted this. This isn't a duplicate. Proof is based on ProofWiki. But I leave out the redundant $a$. ...
1
vote
2answers
1k views

Show that the set of non zero rationals is not a cyclic group under multiplication.

Show that the set of non zero rationals is not a cyclic group under multiplication.I know the set of rationals is not a cyclic group under addition. In an exercise of my book it is given that the ...