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?

$G\cong\Bbb Z_{18}$.  The subgroups are all cyclic, of orders $1,2,3,6,9$ and $18$.  But $H$ has an element of order two ($-1$).  Thus its order is even.  But the order of $H$ is greater than $2$, since it contains $1,-1,-8$.  
Thus it's the subgroup of order $6$, or $G$ itself. 
It is not $G$ though:
Note that we have $|8|=|2^3|=18/(18,3)=18/3=6$, $|-8|=|2^{12}|=18/(12,18)=3$.   And since $G$ is abelian, for any two elements $a,b\in G$, we have $|ab||\operatorname{lcm}(|a|,|b|)$.  Thus the maximum order of an element of $H$ is $6$. 
