I have a following group
$(\mathbb{Z}_{18}^{*}, \cdot)=\{x \in \mathbb{Z}_{18} : gcd(x, 18)=1 \} = \{1, 5, 7, 11, 13, 17\}$
I have just found all the subgroups of $\mathbb{Z}_{18}^{*}$ which are:
$G_{1}=\{1\}$
$G_{2}=\{1, 17\}$
$G_{3}=\{1, 5, 11\}$
$G_{4}=\{1, 7, 13\}$
$G_{5}=\{1, 5, 7, 11, 13, 17\}$
I also know that the group is cyclic if and only if there is exactly one subgroup of each order dividing the order of the main group.
At the same time, a group is cyclic if there exists a generator. One possible generator of this group that I know of is $<5>$.
The question is wheter I have made a mistake somewhere or misunderstood something as the group has a generator which means it should be cyclic, but in the other hand there exists two different subgroups of $ord(3)$, so it should not be cyclic. I would like to know which one it is.
Best Answer
Andry, $G_3$ isn't a subgroup. Note that, in $\mathbb{Z}_{18}$, we have $5^2=25=7\not\in G_3$, what is contradiction. Do you see?
Sorry, I didn't see the comments before. What should I do? Vote to delete this post?