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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?

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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:

$$\{1\}\,,\,\,\{1,9\}\,,\,C_4$$

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

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+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
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Already done :-) –  amWhy Dec 30 '12 at 13:28
    
what c4 means??what elements it contains? –  baaa12 Dec 30 '12 at 13:28
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@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
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Observe your "groups":
A set cannot be a subgroup unless it also contains the identity element of the original group!

Recall:

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

$H$ is closed under $*$,

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

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

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