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
2
$\begingroup$
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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0...and as a punishment you're going to upvote my answer now! :) – 2012-12-30
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1Already done :-) – 2012-12-30
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0what c4 means??what elements it contains? – 2012-12-30
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0baaa12 $C_4$ often is used to represent any cyclic group of order 4, all of which are isomorphic to $(\mathbb{Z}_4, +)$. – 2012-12-30
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0@amWhy I guess i dont understand the calculation why 1,9 is subgroup? – 2012-12-30
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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 $h \in H, h^{-1} \in H$).