Find all subgroups of $(U(Z_{15}),*).$ Need to find all subgroups of $(U(\mathbb{Z}_{15}),\cdot)$. So $U(\mathbb{Z}_{15})$ contains integers $x < 15$ so that $x$ and $15$ are coprime. So
$$
U(\mathbb{Z}_{15}) = \{\overline{1},\overline{2},\overline{4},\overline{7},\overline{8},\overline{11},\overline{13},\overline{14}\}
$$
I know that $\mathbb{Z}_{15}$ is not cyclic. Right now I have used every element of the set to generate a subgroup and got the following $7$ right now:
$$
\{\overline{1}\},\{\overline{1},\overline{4}\},\{\overline{1},\overline{11}\},\{\overline{1},\overline{14}\},\{\overline{1},\overline{2},\overline{4},\overline{8}\},\{\overline{1},\overline{4},\overline{7},\overline{13}\},U(\mathbb{Z}_{15})
$$
How do I know that I have now found all of them? For example $\{\overline{1},\overline{4},\overline{14},\overline{11}\}$ seems to be a good candidate as well but I cannot find it using the generating technique.
 A: The order of the group is $8$, so proper non trivial subgroups can only have orders $2$ and $4$.
You got right the cyclic subgroups; now there can be of noncyclic of order $4$. A noncyclic group of order $4$ is the product of two (cyclic) groups of order $2$ and contains three elements of order $2$, being generated by any two of them. So, take two elements $2$ and consider the subgroup they generate:
$$
\langle\overline{4},\overline{11}\rangle=
\{\overline{1},\overline{4},\overline{11},
\overline{4}\cdot\overline{11}=\overline{14}\}
$$
(no other product is necessary, as you can easily verify).
This fills the list of elements of order $2$, so you're finished.
A: Hint: This group is isomorphic to $G={\mathbf Z}_4\times \mathbf Z_2$, in which the group operation is much easier to understand. Find all nontrivial cyclic subgroups of $G$. Then find all two-generated subgroups of $G$. Then all three-generated subgroups of $G$. Keep going until you stop getting new groups (or, for a slightly more clever approach, until you stop getting new groups of non-prime index in $G$).
