Finding subgroups of $\mathbb{Z}_{13}^*$ 
I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)

My attempt:
$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$ 
Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$ 
$$\begin{align}
&12^1=12\mod 13\\
&12^2=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{12,1\}}$ is a subgroup of order $2$ 
$$\begin{align}
&3^1=3\mod 13\\
&3^2=9\mod 13\\
&3^3=1\mod 13\\
\end{align}$$
$$\begin{align}
&9^1=9\mod 13\\
&9^2=3\mod 13\\
&9^3=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{9,3,1\}}$ is a subgroup of order $3$ 
$$\begin{align}
&5^1=5\mod 13\\
&5^2=12\mod 13\\
&5^3=8\mod 13\\
&5^4=1\mod 13\\
\end{align}$$
$$\begin{align}
&8^1=8\mod 13\\
&8^2=12\mod 13\\
&8^3=5\mod 13\\
&8^4=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{1,5,12,8\}}$ is a subgroup of order $4$ 
$$\begin{align}
&4^1=4\mod 13\\
&4^2=3\mod 13\\
&4^3=12\mod 13\\
&4^4=9\mod 13\\
&4^5=10\mod 13\\
&4^6=1\mod 13\\
\end{align}$$
$$\begin{align}
&10^1=10\mod 13\\
&10^2=9\mod 13\\
&10^3=12\mod 13\\
&10^4=3\mod 13\\
&10^5=4\mod 13\\
&10^6=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{4,3,12,9,10,1\}}$ is a subgroup of order $6$ 

Is it correct? is there any easier method?


Since $2$ is a generator of $\Bbb{Z}_{13}^*$ you have that the subgroups are exactly
$$\{ \langle 2^n \rangle : n \mbox{ divides } 12\}$$
i.e.
$$\langle 2 \rangle \\ \langle 2^2 \rangle = \langle 4 \rangle \\ \langle 2^3 \rangle = \langle 8 \rangle \\ \langle 2^4 \rangle = \langle 3 \rangle \\ \langle 2^6 \rangle = \langle -1 \rangle \\ \langle 2^{12} \rangle = \langle 1 \rangle \\$$
