The subgroups of a cyclic group We've got $G=U(\mathbb Z/(27)\mathbb Z)=\langle 2 \rangle$ a cyclic group, and $H=\langle -8, -1 \rangle$ a subgroup of $G$. I've calculated all the subgroups of $G$.
Now I have to indentify $H$ with a subgroup of $G$, without calculating all the elements of $H$.

So I think that I can see clearly that $H$ is equal to the subgroup $\langle 8 \rangle =\{8,10,-1,-8,-10,1\}$, but as the problem says that I can't calculate all the elements of H to solve this problem, I don't know how can I justify that H is equal to $\langle 8 \rangle$. How can I do it?

We have $|G|=27-9=18$, hence $2^9 = -1$. 
Now $H=\langle -8,-1 \rangle = \langle -2^3, 2^9\rangle  = \langle 2^9\cdot 2^3, 2^9\rangle = \langle 2^{12},2^9\rangle$.
Now thanks to the Bezout Lemma we have $H=\langle 2^{12},2^9\rangle = \langle 2^3\rangle = \langle 8\rangle$.
