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Cyclic groups

The groups generated by one element: their generators counted by Euler's function, one subgroup for each divisor of the order, and one such group of each size.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to decide whether a group is cyclic by looking for an element of full order, list the generators of a cyclic group and count them with Euler's function, list its subgroups and their sizes from the divisors of the order, say why every group of prime order is cyclic, distinguish cyclic from abelian with the Klein four-group as the standard counterexample, and prove that every subgroup of a cyclic group is cyclic.

2. One element, and everything it reaches

Every element $g$ of every group generates a subgroup $\langle g \rangle$, whose size is the order of $g$. Sometimes that subgroup is the whole group — and when it is, the group is as simple as a group can be. This lesson is about that case, and it is the only family of groups this course completely classifies.

3. Cyclic, generator, isomorphic

A group is cyclic when $G = \langle g \rangle$ for some $g$, and any such $g$ is a generator. Two groups are isomorphic, written $G \cong K$, when there is a bijection between them that respects the operations — the same group wearing different labels. Euler's function $\phi(n)$ counts the integers from $1$ to $n$ coprime to $n$.

4. The one family that is completely understood

$G$ is cyclic when every element is a power of one element $g$. There are exactly two kinds.

Infinite. $g$ has infinite order and $G \cong \mathbb{Z}$. The generators are $g$ and $g^{-1}$, and no others.

Finite of order $n$. $G = \{e, g, \ldots, g^{n-1}\}$ and $G \cong \mathbb{Z}_n$. Four facts follow.

1. Cyclic groups are abelian, since $g^{i}g^{j} = g^{i+j} = g^{j}g^{i}$.
2. The generators are the $g^{k}$ with $\gcd(k, n) = 1$, so there are $\phi(n)$ of them. In $\mathbb{Z}_n$ they are the residues coprime to $n$.
3. One subgroup for each divisor. For each $d \mid n$ there is exactly one subgroup of order $d$, namely $\langle g^{n/d} \rangle$, and there are no others. So the subgroups are counted by the divisors of $n$.
4. Every subgroup is cyclic, and so is every quotient.

The classification is the strong statement: two cyclic groups of the same size are isomorphic. So there is one cyclic group of each order, and $\mathbb{Z}_{12}$, the rotations of a regular twelve-sided figure and the twelfth roots of unity under multiplication are three names for it.

The converse of fact 1 fails, and the failure matters: the Klein four-group is abelian and not cyclic.

An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.

5. Which groups turn out to be cyclic

The classification is easy to state and its usefulness is in recognising cases of it, which is not always easy.

Every group of prime order is cyclic. Take any $g \ne e$; its order divides $p$ and is not $1$, so it is $p$, and $\langle g \rangle$ is everything. So there is exactly one group of each prime order — the strongest classification result available this early, and it comes out of Lagrange in one line.

The units modulo $n$ are cyclic for some $n$ and not others. $U(9)$ is cyclic of order $6$, generated by $2$; $U(8)$ is the Klein four-group and is not cyclic. The pattern — cyclic exactly when $n$ is $1, 2, 4$, a power of an odd prime, or twice one — is a theorem of number theory and is not proved here. What matters is that being a group of units does not make a group cyclic, and the only way to decide is to look at the orders.

The additive group of the integers is cyclic, generated by $1$ or by $-1$. Its subgroups $n\mathbb{Z}$ are all cyclic, all infinite except $\{0\}$, and all isomorphic to $\mathbb{Z}$ itself — an infinite group can be isomorphic to a proper subgroup of itself, which no finite group can.

A product need not be. $\mathbb{Z}_2 \times \mathbb{Z}_2$ is the Klein four-group and is not cyclic, while $\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6$ is. The rule is that $\mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic exactly when $m$ and $n$ are coprime, which is the Chinese remainder theorem in group clothing.

6. What being cyclic does and does not mean

Confusing cyclic with abelian. Cyclic implies abelian; abelian does not imply cyclic. The Klein four-group is the smallest counterexample and it is worth remembering by name.

Thinking every element generates. Only the $\phi(n)$ elements coprime to $n$ do. In $\mathbb{Z}_{12}$ that is four elements out of twelve.

Thinking the generator is unique. It is unique only when $\phi(n) = 1$, that is for $n = 1$ and $n = 2$. A cyclic group usually has several generators and they are interchangeable.

Expecting several subgroups of the same size. In a cyclic group there is exactly one subgroup of each divisor size. In a general group there can be many — the symmetries of a square have three subgroups of order $4$ — and that difference is one of the ways a group announces it is not cyclic.

Believing a group written multiplicatively cannot be $\mathbb{Z}_n$. Notation is not structure. The fourth roots of unity under multiplication are $\mathbb{Z}_4$.

7. Generators and subgroups of the integers modulo eighteen

1. Generators: the residues coprime to $18$, namely $1, 5, 7, 11, 13, 17$ — and $\phi(18) = 6$, which agrees.

   Six generators out of eighteen elements.

2. Divisors of $18$: $1, 2, 3, 6, 9, 18$, so six subgroups, of exactly those sizes.

   One per divisor.

3. The subgroup of order $6$ is $\langle 3 \rangle = \{0, 3, 6, 9, 12, 15\}$, since $18/6 = 3$.

   Generated by the modulus divided by the size.

8. Two groups of order four, told apart

1. $\mathbb{Z}_4$: the element $1$ has order $4$, so the group is cyclic and has $\phi(4) = 2$ generators, $1$ and $3$.

   An element of full order.

2. $U(8) = \{1, 3, 5, 7\}$: $3^{2} = 5^{2} = 7^{2} = 1$, so every non-identity element has order $2$.

   No element of order $4$.

3. So $U(8)$ is not cyclic, and the two groups are not isomorphic even though both are abelian of order $4$. Both are classified; only one is cyclic.

   Abelian is not enough.

9. Your turn: how many subgroups does a cyclic group of order $20$ have, and how many generators?

1. Subgroups correspond to divisors of $20$: $1, 2, 4, 5, 10, 20$.

   Count the divisors.

2. So six subgroups, of those six sizes, one each.

   One per divisor, no more.

3. Your turn: work this step out. Its working is at the end of the packet.

   Generators: $\phi(20) = \phi(4)\phi(5) = 2 \times 4 = 8$. The eight residues coprime to $20$ are $1, 3, 7, 9, 11, 13, 17, 19$.

10. Guided practice

The integers modulo $35$ under addition. Say how many elements each of these four subgroups has.

11. Guided practice

How many elements generate the whole of the integers modulo $10$ under addition?

Answer:

12. Practice

In the integers modulo $10$ under addition, match each element to the number of elements in the subgroup it generates.

13. Practice

Select every statement that is true of cyclic groups.

This task has no paper form; do it on a device.

14. Practice

Is $U(8) = \{1, 3, 5, 7\}$ under multiplication modulo $8$ cyclic?

• Yes, cyclic
• No, not cyclic

15. Somewhere new

Build the proof that every subgroup of a cyclic group is cyclic.

This task has no paper form; do it on a device.

16. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

17. Test question

The integers modulo $15$ under addition. Say how many elements each of these four subgroups has.

18. What you can do now

You can find the generators and the subgroups of a cyclic group, and decide whether a given group is cyclic. Say in your own words why an abelian group need not be cyclic, and name one that is not. Next: permutations, where the groups stop commuting for good.

Working for the steps left to you

9. Your turn: how many subgroups does a cyclic group of order $20$ have, and how many generators?, step 3