Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say I want to list all the cyclic subgroups of $G$. Let's say $G = \mathbb{Z}^*_{10}$. Then I know all the elements in $G$ are $1, 3, 7, 9$ so all I need know is to find the cyclic subgroups from those elements. As I understand I need to find subgroups so that all elements generate from one element? Then if I'm right the subgroups are $\{1\}, \{3, 9\}, \{7\}, \{9\}$? Is that right?

share|cite|improve this question
up vote 3 down vote accepted

Perhaps it's easier to note that $\,\Bbb Z_{10}^*\cong C_4=$ the cyclic group of order $\,4\,$, so that there are exactly

three subgroups here:


Check that $\,\{3\}\,,\,\{9\}\,$ cannot be subgroups as they don't contain the unit element...

share|cite|improve this answer
+1 Beat me to it! :-/ – amWhy Dec 30 '12 at 13:25
...and as a punishment you're going to upvote my answer now! :) – DonAntonio Dec 30 '12 at 13:27
Already done :-) – amWhy Dec 30 '12 at 13:28
what c4 means??what elements it contains? – baaa12 Dec 30 '12 at 13:28
@baaa12: $\,9^2=81=1\pmod {10}\,$ . In any group with an element $\,x\,$ of order two (an involution), the set $\,\{1,x\}\,$ is a subgroup (of order two, of course) – DonAntonio Dec 30 '12 at 13:41

Observe your "groups":
A set cannot be a subgroup unless it also contains the identity element of the original group!


$H\le (G,*) \iff $:

$H$ is closed under $*$,

The identity of $G$ is IN $H$.

$H$ is closed under inversion. (For all $h \in H, h^{-1} \in H$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.