$H = \{ \frac{7}{6}n + \frac{5}{8}m \mid n, m \in \Bbb{Z} \}$ is a subgroup of $(\Bbb{Q}, +)$. Prove that $H$ is cyclic.

Question

$H = \{ \frac{7}{6}n + \frac{5}{8}m \mid n, m \in \Bbb{Z} \}$ is a subgroup of $(\Bbb{Q}, +)$. Prove that $H$ is cyclic

So from what I see, I need to find that an element from $H$ where $<h^k>$ = $H$. This means that I can use $(\frac{7}{6}n + \frac{5}{8}m)^k$... Can I use the GCD for this?

so I added and got $<\frac{28n + 15m}{24}>$ and since LCM(28,15) = 1 then the linear combination can produce all the integers in Z? Therefore $\frac{1}{24}$ is a generator???

Answer 1

Note how: $$\frac76n+\frac58m=\frac 1{24}(28n+15m)$$ And since $\gcd(28,15)=1$, we can find $n,m\in\mathbb{Z}$ with: $$28n+15m=1$$ Hence $\frac1{24}\in H$. Therefore: $$\langle\frac1{24}\rangle\subseteq H$$ But: $$H=\{\frac76n+\frac58m|n,m\in\mathbb{Z}\}=\{\frac1{24}(28n+16m)|n,m\in\Bbb{Z}\}\subseteq\{\frac{n}{24}|n\in\Bbb{Z}\}=\langle\frac1{24}\rangle$$ We conclude that $H=\langle\frac1{24}\rangle$, so $H$ is cyclic.