Is $\mathbb{Z}_{18}^{*}$ a cyclic group?

Question

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.

Answer 1

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?

Answer 2

Indeed, ${\Bbb Z}_{18}^* = \langle 5\rangle$, since $5^1=5$, $5^2=7$, $5^3=17$, $5^4=13$, $5^5=11$, and $5^6=1$ thanks to Maple.

Answer 3

Indeed, it is a general fact that $\mathbb{Z}_n^*$ is cyclic if and only if $n = 2,4,p^k,2p^k$ where $p$ is an odd prime. This follows from the chinese remainder theorem; a discussion of which can be found in Dummit & Foote or some equivalent text. Since $18 = 2 \times 3^2$, it satisfies the criterion above and hence the group in question is cyclic.

Answer 4

You have already seen it has order 6, and you know it is abelian, since integer multiplication is commutative. All abelian groups of square-free order are cyclic, being a direct product of cyclic groups of coprime order (namely their Sylow subgroups).