I am trying to better understand the cyclic groups and so far understand that a non-trivial subgroup is a subgroup that DOES NOT contain just the element identity alone, i.e. $\langle 6\rangle = \{6\}$ is a TRIVIAL subgroup of $\mathbb{Z}/6\mathbb{Z}$.

However, how would one go about finding the NON-TRIVIAL subgroups? Lets say for example cyclic group $G = (\mathbb{Z}/89\mathbb{Z})^\times$. The factors of $89-1=88$, are $\{1,2,4,8,11,22,44,88\}$ so these are the orders of the elements.

Now I think I found a few NON TRIVIAL subgroups, but I am not sure if they are correct $$ \langle 3\rangle = \{3,9,12,15,18,\ldots\}\\ \langle 4\rangle = \{4,8,12,16,\ldots\}\\ \langle 15\rangle = \{15,30,45,\ldots\} $$

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    $\begingroup$ There's NO NEED to SHOUT :P $\endgroup$
    – Shaun
    Aug 24, 2019 at 14:59
  • $\begingroup$ Any subgroup of any cyclic group is cyclic. Just pick powers of the generator that share a factor with the order of the original group. $\endgroup$
    – Shaun
    Aug 24, 2019 at 15:05
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    $\begingroup$ The question is really unclear. What do you mean by $\langle 6\rangle=\{6\}$? What is $6$ here? An element of which group? $\endgroup$
    – Mark
    Aug 24, 2019 at 15:08
  • $\begingroup$ @Shaun what would be some examples please Sir $\endgroup$ Aug 24, 2019 at 15:09
  • $\begingroup$ See my answer below. It delineates the non-trivial subgroups of cyclic groups. $\endgroup$
    – Shaun
    Aug 24, 2019 at 15:21

2 Answers 2


The subgroups you found are indeed non-trivial. In general the cyclic subgroup $\langle g\rangle$ generated by any non-identity element $g$ of a group $G$ must be non-trivial, since it contains at least $g$. (To enumerate all the elements of a cyclic subgroup of a finite group, just keep taking powers of the generator $g$ until you have the identity in your subgroup, then stop.)

For example, in your group $(\mathbb{Z}/89\mathbb{Z})^\times$, you can take $g=55$ and calculate its powers. We find $g^1 = 55$, $g^2 = 3025$, which $\equiv 88 \pmod{89}$. Then $g^3 = 166375 \equiv 34 \pmod{89}$, and finally $g^4 = 9150625 \equiv 1\pmod{89}$ is the identity. So $\langle 55\rangle = \{55, 88,34,1\}$ is a non-trivial cyclic subgroup of order 4. (Of course, we could have avoided the huge powers by performing multiplication on the "modded" terms.)

  • $\begingroup$ Could you be kind enough to give some examples please? $\endgroup$ Aug 24, 2019 at 15:09
  • $\begingroup$ You can take any group you like to be an example! Then choose an element $g$ that is not the identity, then keep taking powers of it $g, g^2, g^3, \ldots$ until you hit $g^n=e$ where $n$ is the order of $g$ and $e$ is the identity. Then $\{g, g^2, \ldots,e\}$ is a cyclic subgroup that is not trivial. $\endgroup$
    – marcelgoh
    Aug 24, 2019 at 15:12
  • $\begingroup$ Sir thank you Sir, I don't suppose you could be so kind enough to do me some examples please? God Bless $\endgroup$ Aug 24, 2019 at 15:14
  • $\begingroup$ Okay, take $G = S_3$, the permutation group on 3 letters, and $g = (123)$. This generates the subgroup $\{(123), (123)^2, (123)^3\} = \{(123), (132), e\}$. $\endgroup$
    – marcelgoh
    Aug 24, 2019 at 15:22
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    $\begingroup$ @Shaun, you're right, it only works in finite groups. I'll amend my answer. $\endgroup$
    – marcelgoh
    Aug 24, 2019 at 15:36

A presentation of a (finite) cyclic group $\Bbb Z_n$ for $n\in \Bbb N$ is

$$\langle a\mid a^n\rangle.$$

Suppose $n=mk$ for some non-trivial $m, k\in\Bbb N$. Let $b=a^m$. Then $$\langle b\mid b^k\rangle$$

defines a group isomorphic to a subgroup of $\Bbb Z_n$ of order $k$.

For an infinite cyclic group, say, $(\Bbb Z, +)$, then the subgroup given by $(\nu \Bbb Z, +)$ for some $\nu\in\Bbb N\setminus\{1\}$ is non-trivial. (Why?) Here $\nu\Bbb Z=\{\nu z\mid z\in\Bbb Z\}$.

  • $\begingroup$ Thank you Sir, I don't suppose you can do a few examples for my case please? I am self taught student and willing to learn new things everyday. would really mean a lot if you could kindly help me $\endgroup$ Aug 24, 2019 at 15:23
  • $\begingroup$ Sure: $\Bbb Z/3\Bbb Z$ is isomorphic to the subgroup $\langle [2]_6\rangle=(\{[0]_6, [2]_6, [4]_6\}, +_6)$ of $\Bbb Z/6\Bbb Z$, generated by $[2]_6$, because $\gcd(6, 2)\neq 1$. $\endgroup$
    – Shaun
    Aug 24, 2019 at 15:31
  • $\begingroup$ Does that help, @TomazWiszvortiox? $\endgroup$
    – Shaun
    Aug 24, 2019 at 15:55

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