# Are cyclic groups always abelian? [closed]

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?

• Have you tried proving this yourself? Where do you get stuck? The "explanation" is that an element always commutes with powers of itself. – Qiaochu Yuan Jun 12 '11 at 19:27
• In fact, not only is every cyclic group abelian, every quasicylic group is always abelian. (A group is quasicyclic if given any $x,y\in G$, there exists $g\in G$ such that $x$ and $y$ both lie in the cyclic subgroup generated by $g$). The Prufer $p$-group and the rationals (under addition) are examples of quasicylic groups that are not cyclic. – Arturo Magidin Jun 12 '11 at 20:02

Yes, a cyclic group is abelian. Here is why.

A cyclic group is generated by one generator, let's call this $g$. Now if $a = g^m$ and $b = g^n$ are two elements of the group, then $ab = g^{m}g^n = g^{n}g^m = ba$ (since $g$ commutes with itself).

• The main point is of course that $g^m g^n = g^{m+n} = g^{n+m} = g^n g^m$. – lhf Jun 13 '11 at 11:28

The following exercises (in order of increasing difficulty) will strengthen your understanding of the ideas presented above:

Exercise 1:

Give an example of an abelian group that is not cyclic.

Exercise 2:

Let $G$ be a group such that every proper subgroup of $G$ is cyclic. Is $G$ necessarily cyclic?

Exercise 3:

Let $G$ be a finite group with a unique maximal subgroup. Prove that $G$ is cyclic. (Hint: let $M$ be the unique maximal subgroup of $G$. Recall that the definition of "maximal subgroup" also requires that the subgroup be proper. Hence we can find $x\in G,x\not\in M$. Prove that $x$ generates $G$.)

Exercise 4:

Prove that a non-trivial abelian simple group is cyclic of prime order. (Recall that a group $G$ is said to be simple if it does not have any proper non-trivial normal subgroups.) (Hint: which subgroups of an abelian group are normal? What can you say about non-trivial groups with exactly two subgroups?)

Exercise 5:

Let $G$ be a finite group such that the quotient group $G/Z(G)$ is cyclic. Prove that $G$ is abelian, that is, prove that $G=Z(G)$. (Recall that $Z(G)$ is the center of $G$; namely, it is the set of all elements in $G$ that commute with every element of $G$.)

Challenging Exercises:

Exercise A:

Prove that an abelian group of order $6$ is cyclic. In fact, prove that there are exactly two isomorphism classes of groups of order $6$. Also, give representatives of these isomorphism classes.

Exercise B:

Prove that a group of order $p^2$ is abelian if $p$ is prime. (Hint: first prove (or use if you cannot prove) that if $G$ is such a group, then $Z(G)\neq 1$, that is, the center of $G$ is non-trivial. Then consider a case-by-case argument based on Exercise 5 above.)

Exercise C:

Prove that every group of order $35$ is cyclic. (Hint: first prove (or use if you cannot prove) that if $G$ is such a group, then $G$ has exactly one normal subgroup of order $5$ and of order $7$.)

• These are excellent exercises, but hardly an answer to the question. – lhf Jun 13 '11 at 11:26
• Dear Ihf, you are of course correct. However, since the question is already answered by $4$ other users, I thought it would be better to supplement these answers with some additional related results (given as exercises). The nature of the question suggests that the definition of a cyclic group is fairly new to the OP and these exercises should improve his understanding. Also, some exercises are at least a little relevant; most of the exercises are based on the relation between "abelian and "cyclic". However, if the community feels that this answer is inappropriate, then I am happy to delete. – Amitesh Datta Jun 13 '11 at 12:05
• This is surely not the place to suggest new exercises to the user – Ishan Srivastava Nov 23 '18 at 5:08
• @Amitesh Datta thanks great help – Eden Hazard Jun 4 '19 at 21:43

Yes, all cyclic groups are abelian. Here's a little more detail that helps make it explicit as to "why" all cyclic groups are abelian (i.e. commutative).

Let $G$ be a cyclic group and $g$ be a generator of $G$. Let $a,b \in G$. Since $g$ is a generator of $G$, all elements in $G$ can be expressed as integral powers of $g$ (or in the case the group operation on $G$ is additive, all the elements of $G$ can be expressed integral multiples of $g$). Then there exist $x,y \in \mathbb {Z}$ such that $a=g^x$ and $b=g^y$ (or $a = xg$ and $b = yg.$). Since $ab=g^xg^y=g^{x+y}=g^{y+x}=g^yg^x=ba$, (or $a+b = xg+ yg= yg + xg = b + a$) it follows that $G$ is abelian.

This proof helps show what Qia means in the first comment above by "an element always commutes with powers of itself." The key with cyclic groups is that all elements of a given cyclic group can be expressed in terms of one element in the group: as an integral power (or multiple) of the group's generator. (If a group $G$ has more than one generator, each element in $G$ can be expressed in terms of each of the group's generators).

The easy explanation is that the exponents are integers and integer addition is commutative: $g^m g^n = g^{m+n} = g^{n+m} = g^n g^m$.

The harder explanation is that all cyclic groups are homomorphic images of $\mathbb Z$, which is abelian.

• Infinite cyclic groups are isomorphic to $\mathbb Z$ (under addition). All finite cyclic groups are isomorphic to $\mathbb Z_n$ (under addition modulo $n$). – Namaste Jun 12 '11 at 20:13

Let $G$ be a group that is generated by a single element $g$. Let us define $C_g$ as the group of all elements $x \in G$ such that $xg = gx$. In other words: $$C_g = \{ x\in G | xg = gx \}$$

I'll leave proving that $C_g$ is a subgroup as an exercise.

Now, since $g$ satisfies $gg=gg$, then $g \in C_g$.

Convince yourself that the fact that $g \in C_g$ gives $C_g = G$, since $g$ generates $G$. This gives that $g$ is commutative to every other element in $G$.

Let us look at $Z(G) = \{x \in G | xy = yx \forall y \in G\}$, ie the group of elements that are commutative to all other elements in $G$.

Again, since every other element is commutative to $g$ as we saw above (from the fact that $C_g = G$), we get that $g \in Z(G)$ and again since $g \in Z(G)$ we get that $Z(G) = G$.

This final result says that every element in $G$ is commutative to every other element in $G$, meaning that $G$ is abelian.

Yes, cyclic groups are Abelian.

For instance, consider G such that

$$\begin{array}{|c| c c c} & g^{0} & g^{1} & g^{2}\\\hline g^{0} & g^{0} & g^{1} & g^{2}\\ g^{1} & g^{1} & g^{2} & g^{0}\\ g^{2} & g^{2} & g^{0} & g^{1} \end{array}$$

Where $g^{0}$ is the identity.

Just thought I'd add a visual depiction since the textual explanations were already covered.

Edited re Elliott's comment

• So this is $\mathbb Z/3$? – Rasmus Jun 12 '11 at 19:58
• This is a valid example, but giving a single example doesn't explain why we'd expect cyclic groups to be abelian in general. – Elliott Jun 12 '11 at 20:03
• Oh, I certainly didn't mean to explain the generality. There are many much better explanations with higher votes. I was actually going to make a comment about homomorphic images of $\mathbb{Z}$, but that'd already been mentioned. I thought that a simple image would serve as a helpful example, if rather lacking in abstraction and rigor. – Jack Henahan Jun 12 '11 at 20:16
• Sounds good. And in fact, having this picture is great because it shows that each row (or column) is the previous row (or column) shifted by 1. What's more is that if we associate $e,a,b$ with $g^0, g^1, g^2$, then we can see the "addition table" structure come out. – Elliott Jun 12 '11 at 20:25
• @Elliott That is actually much clearer. I only wish I'd thought of it. :P – Jack Henahan Jun 12 '11 at 20:35