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?
Cyclic subgroups of finite groups
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$\begingroup$
abstract-algebra
group-theory
cyclic-groups
2 Answers
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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@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) – 2012-12-30
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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 $h \in H, h^{-1} \in H$).