List all the cyclic subgroups of $\langle\mathbb Z_{10}, +\rangle$ I am trying to get the hang of it, so my work is very belabored. Please, see if any of that is correct. 
Counterexamples: 
$\mathbb Z_2 = \{0, 1\}$ is not subgroup since $1 + 1 = 2$ is outside $\mathbb Z_2.$
$\mathbb Z_3 = \{0, 1, 2\}$ is not a subgroup since $3 + 1 = 4$ is outside $\mathbb Z_5.$
$\mathbb Z_4 = \{0, 1, 2, 3\}$ is not a subgroup since $3 + 1 = 4$ is outside $\mathbb Z_4.$
$\mathbb Z_5 = \{0, 1, 2, 3, 4\}$ is not a subgroup since $4 + 1 = 5$ is outside $\mathbb Z_3.$
$\mathbb Z_6 = \{0, 1, 2, 3, 4, 5\}$ is not a subgroup since $5 + 1 = 6$ is outside $\mathbb Z_6.$
$\mathbb Z_7 = \{0, 1, 2, 3, 4, 5, 6\}$ is not a subgroup since $6 + 2 = 8$ is outside $\mathbb Z_7.$
$\mathbb Z_8 = \{0, 1, 2, 3, 4, 5, 6, 7\}$ is not a subgroup since $5 + 3 = 8$ is outside $\mathbb Z_8.$
$\mathbb Z_9 = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$ is not a subgroup since $7 + 2 = 9$ is outside $\mathbb Z_9.$

$\mathbb Z_1 = \{0\}$.
$0 + 0 = 0$, so $0 \in \mathbb Z_1$ and $0$ is the identity and its own inverse, so $\mathbb Z_1$ is a subgroup of $\mathbb Z_1$.
$\mathbb Z_{10} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$
Let $c \neq 0 \in \mathbb Z_{10}.$ Then $c + 0 = c, \forall c \in \mathbb Z_{10}.$ So, $0$ is the identity. $c + d = 0 \to d = -c,$ so $-c$ is the inverse. The sum of any two elements in $Z_{10}$ is still in $Z_{10}.$ So $\mathbb Z_{10}$ is a subgroup of $\mathbb Z_{10}$. 

Your attempt can't be successful. Since any subgroup must contain $0$, you have to consider all subsets of $\mathbb{Z}_{10}$ that contain $0$ and there are $2^9=512$ of them.
You have a basic information: you want to list all cyclic subgroups, so they must have a generator by assumption. Hence you just consider
$$
\langle 0\rangle=\{0\}\\
\langle 1\rangle=\mathbb{Z}_{10}\\
\langle 2\rangle=\{0,2,4,6,8\}\\
\langle 3\rangle=\{0,3,6,9,2,5,8,1,4,7\}=\mathbb{Z}_{10}\\
\dots
$$
(just start from one element, put in $0$, then keep adding the chosen element until you get back to $0$).
You'll discover that there are just four cyclic subgroups. Indeed, a basic theorem on groups tells you that the subgroups of $\mathbb{Z}_{10}$ are in one to one correspondence with the subgroups of $\mathbb{Z}$ containing $10\mathbb{Z}$, so $10\mathbb{Z}$, $2\mathbb{Z}$, $5\mathbb{Z}$, $\mathbb{Z}$. And, since these are cyclic, also their images in $\mathbb{Z}_{10}$ are cyclic.
Another important thing to note: $\mathbb{Z}_n$ is not a subgroup of $\mathbb{Z}_m$ if $m\ne n$.
