Find all subgroups of group Given multiplicative group of integers modulo 13, namely $\mathbb{Z}_{13}^*$, find all subgroups of this group. I need to prove that this group is cyclic. Also, as $|\mathbb{Z}_{13}^*| = 12$, I know that, if $H$ is a subgroup of $\mathbb{Z}$, then $|H|$ is one of ${1, 2, 3, 4, 6, 12}$. And I got stuck here. No computing.
 A: If $G=Z^*_{13}$ is cyclic, then every subgroup of $G$ is also cyclic, and so must be generated by a single element.  You can find all the subgroups by considering the 12 groups of the form $\langle s \rangle$ for $s\in G$; some of these 12 will be the same, and you need only cross the duplicates off your list of 12.
While considering these 12 subgroups, you will find that one of them, say $\langle i\rangle$, is equal to $G$; this will prove that $G$ is cyclic, and tell you that $G$ is generated by $i$.
A: that group is the multiplicative group of the field $\mathbb Z_{13}$, the multiplicative group of any finite field is cyclic. For a proof see here.
All you have to do is find a generator (primitive root) and convert the subgroups of $\mathbb Z_{12}$ to those of the group you want by computing the powers of the primitive root.
The subgroups of the group $g,g^2,g^3\dots g^{12}=e$ are those generated by $g^k$ where $k$ divides $12$.
So you want $\langle g \rangle,\langle g^2 \rangle,\langle g^3 \rangle,\langle g^4\rangle, \langle g^6 \rangle, \langle g^{12} \rangle$
