2
$\begingroup$

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?

2 Answers 2

3

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...

  • 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
3

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 $h \in H, h^{-1} \in H$).