<s>[INST] 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?

 [/INST] Your solution is right. As, I  am noting that(may be i'm wrong), you are applying that "A cyclic subgroup of order $4$ must contain $2$ elements of order $4$ and  $1$ element of order $2$, and you searching those elements and listing them. You can reduce your calculation by searching one element of each order, and then you can generate your required subgroups, e.g. $5$ is element of order $4$ so, 
$$<5>=\{1,5,8,12\}$$ is subgroup of order $4$
</s>