For group $\mathbb{Z_{18}^*}$, how do I find all subgroups

Question

In my textbook, there is a cyclic Group $G=\mathbb{Z_{18}^*}$ which has the elements
$$\{1,5,7,11,13,17\}$$

And its subgroups are $U_1 = \{1\}$, $U_2 = \{1,17\}$ and $U_3 = \{1,7,13\}$

How did they come to these subgroups? I tried to use the Definition of subgroups, but could not come to these subgroups? how do I come to this result? why is $5$ being omitted?

Start with any element, and generate a subgroup. All subgroups must arise this way since the group is cyclic. — Prahlad Vaidyanathan, Sep 29 '13 at 12:31
Convince yourself that any subgroup that contains 5 must be the entire group. (So $G$ itself should also be listed as a subgroup of $G$.) — Gerry Myerson, Sep 29 '13 at 12:33
Start with any element $a$, and write down ${a,a^2,a^3,\ldots}$ until you start repeating (ie. stop when you hit the identity). Note that all multiplication is mod$(18)$ — Prahlad Vaidyanathan, Sep 29 '13 at 12:38
@PrahladVaidyanathan, i will go it for each element of G, right? — doniyor, Sep 29 '13 at 12:41

score 6 · Accepted Answer · answered Sep 29 '13 at 12:58

We can verify that $$\langle 1 \rangle=\{1\},$$ $$\langle 5 \rangle=\{1,5,7,17,13,11\}=G,$$ $$\langle 7 \rangle=\{1,7,13\},$$ $$\langle 11 \rangle=\{1,11,13,17,7,5,1\}=G,$$ $$\langle 13 \rangle=\{13,7,1\}=\langle 7 \rangle,$$ $$\langle 17 \rangle=\langle -1 \rangle=\{1,17\}.$$

This gives the four subgroups $\{1\}$, $\langle 7 \rangle$, $\langle -1 \rangle$, and $G$. We could argue that $G$ is cyclic, and thus any subgroup of $G$ is also cyclic, and thus this list is complete.

Alternatively, we see that any subgroup containing $5$ or $11$ must be $G$ itself. Therefore, any subgroup without $5$ nor $11$ must (a) contain $1$, (b) be a subset of $\{1,7,13,17\}$ (c) contain either $1$, $2$ or $3$ elements, by Lagrange's Theorem.

If it contains $1$ element, the subgroup must be $\{1\}$.
If it contains $2$ element, the subgroup must be $\langle 17 \rangle$, since the elements $7$ and $13$ have order $3$.
If it contains $3$ element, it cannot contain $17$ since it has order $2$, and a group of order $3$ cannot contain a subgroup of order $2$, by Lagrange's Theorem. This leaves only $\langle 7 \rangle$.

So the four subgroups identified are all of the subgroups.

It looks like $\langle 5 \rangle$ is omitted since it's not a proper subgroup of $G$ (i.e., it equals $G$ itself).

Rebecca, just a small question: you meant this, right? $\langle 5 \rangle={1=5^0(mod 18),5=5^1(mod 18),7=5^2(mod 18),17=5^3(mod 18),13=5^4(mod 18),11=5^5(mod 18)}$ is this how i find them right? — doniyor, Oct 05 '13 at 10:17

score 2 · Answer 2 · answered Oct 16 '13 at 10:12

Note that $$\mathbb Z_{18}^*\cong\mathbb Z_6=\{0,1,2,3,4,5\}$$ and $$\langle 1\rangle=\langle 5\rangle=\mathbb Z_6$$ The following codes in GAP, give you the desired subgroups:

gap> LoadPackage("sonata");;
     Z_18:=Units(Integers mod 18);;
     e:=Subgroups(Z_18);;
     for i in [1..Size(e)] do Print(e[i],"~",StructureDescription(e[i]),"\n"); od;

 Group( ... )~1
 Group( [ ZmodnZObj( 17, 18 ) ] )~C2
 Group( [ ZmodnZObj( 13, 18 ) ] )~C3
 Group( [ ZmodnZObj( 11, 18 ) ] )~C6

For group $\mathbb{Z_{18}^*}$, how do I find all subgroups

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