Groups, subgroups and quotient groups

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(Q,+) abelian group, (Z,+) his subgroup and Q/Z a quotient group.

  1. Find all subgroups of <$\frac{\hat1}{6}$>,$\frac{\hat1}{6}$ $\in$ Q/Z.

The elements of <$\frac{\hat1}{6}$> are { $\frac{\hat1}{6}$ $\frac{\hat2}{6}$ $\frac{\hat3}{6}$ $\frac{\hat4}{6}$ $\frac{\hat5}{6}$ }. $\frac{\hat6}{6}$ is 1 and $\frac{\hat7}{6}$ is 1+$\frac{\hat1}{6}$ so the order or this group would be 6. Then I'm not really sure what it's subgroups would be. If I take a subset of { $\frac{\hat1}{6}$ $\frac{\hat2}{6}$ $\frac{\hat3}{6}$ $\frac{\hat4}{6}$ $\frac{\hat5}{6}$ } then by summing the elements I'll get one that isn't in the subset so it wouldn't be a subgroup. So are { $\frac{\hat1}{6}$ $\frac{\hat2}{6}$ $\frac{\hat3}{6}$ $\frac{\hat4}{6}$ $\frac{\hat5}{6}$ } and {$\emptyset$} it's only subrougps?

  1. Is the subgroup <$\frac{\hat5}{7}$,$\frac{\hat8}{9}$> cyclic? If it is find a generator for it.

Here, I didn't understand how the elements of this subgroup should look like... {($\frac{\hat(5+5)}{7}$,$\frac{\hat(8+8)}{9}$)($\frac{\hat(5+5+5)}{7}$,$\frac{\hat(8+8+8)}{9}$)...} I know there might be some theory to explain this but it's really hard for me to understand from abstract theory so this example would help me a lot.

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  1. Since $\{\emptyset\}$ isn't even a subgroup of $\left\langle\hat{\frac16}\right\rangle$, it cannot possibly be a subgroup. Besides,$$\left\langle\hat{\frac16}\right\rangle=\left\{\hat{\frac06},\hat{\frac16},\hat{\frac26},\hat{\frac36},\hat{\frac46},\hat{\frac56}\right\}$$Note that $\left\langle\hat{\frac16}\right\rangle$ is a cyclic group of order $6$. Therefore, its subgroups, besides the whole group, are $\left\{\hat{\frac06}\right\}$, $\left\{\hat{\frac06},\hat{\frac26},\hat{\frac46}\right\}$ and $\left\{\hat{\frac06},\hat{\frac36}\right\}$.
  2. Yes, it is cyclic. Since $\frac57+\frac89=\frac{38}{63}+1$, then $\hat{\frac57}+\hat{\frac89}=\widehat{\frac{38}{63}}$. It easy to see that$$\left\langle\hat{\frac57},\hat{\frac89}\right\rangle=\left\langle\widehat{\frac{38}{63}}\right\rangle.$$