# Cyclic Groups, inverses, rings and subgroups

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 $$\langle 3\rangle = \{3,9,12,15,18,\ldots\}\\ \langle 4\rangle = \{4,8,12,16,\ldots\}\\ \langle 15\rangle = \{15,30,45,\ldots\}$$

• There's NO NEED to SHOUT :P – Shaun Aug 24 at 14:59
• 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. – Shaun Aug 24 at 15:05
• The question is really unclear. What do you mean by $\langle 6\rangle=\{6\}$? What is $6$ here? An element of which group? – Mark Aug 24 at 15:08
• @Shaun what would be some examples please Sir – Tomaz Wiszvortiox Aug 24 at 15:09
• See my answer below. It delineates the non-trivial subgroups of cyclic groups. – Shaun Aug 24 at 15:21

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.)

• Could you be kind enough to give some examples please? – Tomaz Wiszvortiox Aug 24 at 15:09
• 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. – marcelgoh Aug 24 at 15:12
• Sir thank you Sir, I don't suppose you could be so kind enough to do me some examples please? God Bless – Tomaz Wiszvortiox Aug 24 at 15:14
• Okay, take $G = S_3$, the permutation group on 3 letters, and $g = (123)$. This generates the subgroup $\{(123), (123)^2, (123)^3\} = \{(123), (132), e\}$. – marcelgoh Aug 24 at 15:22
• @Shaun, you're right, it only works in finite groups. I'll amend my answer. – marcelgoh Aug 24 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 $$\nu\Bbb Z=\{\nu z\mid z\in\Bbb Z\}$$.

• 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 – Tomaz Wiszvortiox Aug 24 at 15:23
• Sure: $\Bbb Z/3\Bbb Z$ is isomorphic to the subgroup $\langle _6\rangle=(\{_6, _6, _6\}, +_6)$ of $\Bbb Z/6\Bbb Z$, generated by $_6$, because $\gcd(6, 2)\neq 1$. – Shaun Aug 24 at 15:31
• Does that help, @TomazWiszvortiox? – Shaun Aug 24 at 15:55