# How to show a group is cyclic?

One question asking if $\mathbb{Z}^*_{21}$ is cyclic.

I know that the cyclic group must have a generator which can generate all of the elements within the group.

But does this kind of question requires me to exhaustively find out a generator? Or is there any more efficient method to quickly determine if a group is a cyclic group?

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I think it is a good idea if you are learning these things for the first time to exhaustively check the elements, but think about ways to cut down on your work, e.g. the possible orders are factors of the size of the group, which is 12, so if an element does not have 6th power congruent to 1 then it must have order 12. Do you know any algebraic properties of cyclic groups? – KCd Jun 18 '14 at 12:26
As your group is obviously finite then this problem is soluble in the way you suggest. However, there was a classic "fun" problem from the 60s which asked to prove that the following group was cyclic of order 11. $$\langle x_0, x_1, x_2, x_3, x_4; x_ix_{i+1\pmod5}=x_{i+2\pmod5}\rangle$$ It was posed by John Conway, and it took the best part of three years for someone to find a solution. – user1729 Jun 18 '14 at 12:29
@KCd Thank you for your answer. I am not an expert of group theory, the only thing I know is about some properties with group: associative, identity and inverse including finding inverse. I also know that the order for this group is phi(21). But I am not sure about the algebraic properties of a cyclic group. – Yang Xia Jun 18 '14 at 12:42
@user1729PhD Thank you:) – Yang Xia Jun 18 '14 at 12:48
I should have said - you can find more details, and a proof, in the book "presentations of groups" by D.L.Johnson. The question first appeared in American Mathematics Monthly, or something similar. Such groups are called cyclically presented, and they are an active research area. If we were working with $n>11$ generators and the relators were modulo $n$ then the groups would be infinite...(I think $n>11$ works. Certainly, $n>20$ works.) – user1729 Jun 18 '14 at 13:01

If an abelian group has elements of order $m$ and $n$, then it also has an element of order $lcm(m,n)$, so that is another potential way to cut down on the work you have to do. So, for example, if you were to find an element of order $4$ and one of order $3$ in your group, you would know that there would have to be an element of order $12$, so it would be cyclic.

It turns out that there is an explicit characterization of $\mathbb{Z}_n^{\times}$ that depends on the factorization of $n$; in your case this becomes $\mathbb{Z}_{21}^{\times} \cong \mathbb{Z}_3^{\times} \times \mathbb{Z}_7^{\times}$, so once you know this theorem, showing whether this group is cyclic or not becomes pretty easy.

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Thanks. Changed. – rogerl Jun 19 '14 at 2:09

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.

However, $Z^*_{21}$ is a rather small group, so you can easily check all elements for generators.

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In practice, to show $(\mathbf Z/n\mathbf Z)^\times$ is not cyclic you can look for a "fake" square root of $1$, i.e., a solution to $a^2 \equiv 1 \bmod n$ with $a \not\equiv \pm 1 \bmod n$. Then there are at least two subgroups of order $2$, so this group is not cyclic.

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I believe this is the same as what KCd said, but I will be more specific. $Z^*_{21}$ contains two subgroups of order 2, namely $<8>$ and $<13>$. However, for $Z^*_{21}$ to be cyclic, it must have only one subgroup of order 2. This fact comes from the fundamental theorem of cyclic groups:

Every subgroup of a cyclic group is cyclic. Moreover, if $|a| = n$, then the order of any subgroup of $<a>$ is a divisor of $n$; and, for each positive divisor $k$ of $n$, the group $<a>$ has exactly one subgroup of order k–namely, $<a ^{n/k}>$. http://www.cs.earlham.edu/~seth/class/math420/Portfolio.pdf

In the context of your question, we apply this theorem by supposing $Z^*_{21}$ is cyclic. Then let $a\in{Z^*_{21}}$ such that $<a>=Z^*_{21}$. Thus $|a|=12$. Since 2 divides 12, by the fundamental theorem of cyclic groups, $<a>$ has exactly one subgroup of order 2. However we know that $Z^*_{21}$ possesses two subgroups of order 2. So, $Z^*_{21}$ is not cyclic.

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