Let $a = 2/3$ and $b=5/7$ and let $G = \langle a,b \rangle \subset (\Bbb Q, +)$. Show that $G$ is cyclic and generated by $1/21$. 
Let $a = 2/3$ and $b=5/7$ and let $G = \langle a,b \rangle \subset (\Bbb Q, +)$. Show that $G$ is cyclic and generated by $1/21$.

I compute $\gcd(a,b) = \gcd(2/3,5/7) = \frac{\gcd(2,5)}{\text{lcm}(3,7)} = 1/21$ which implies that there exists $x,y \in \Bbb Z$ such that $$x(2/5)+y(3/7) = 1/21$$
So to show that $\langle 2/3,5/7 \rangle = \langle 1/21 \rangle $ for for an element  in $\langle 2/3,5/7 \rangle$ I have that $x \in \langle 2/3,5/7 \rangle \iff x = n(2/3) + k(5/7)$, but since Bezout there exists $s,t \in \Bbb Z$ such that $x = s(2/3) + t(5/7) = 1/21$ thus $x \in \langle 1/21 \rangle$.
Now if $y \in \langle 1/21 \rangle$, then $y = n/21$. For $n = 29$ I have that $y = 29/21 = 2/3 + 5/7 \in \langle 2/3, 5/7 \rangle$.
Is it true always that if $\gcd(a,b) = k$, then $\langle a,b \rangle = \langle k \rangle$?
 A: If $a,b\in\mathbb{Q}$ then the $gcd(a,b)$ that you wrote is not well defined.
For example: $1/21=gcd(2/3,5/7)=gcd(4/6,5/7)=1/42$
In order to show that $G\cong \langle1/21\rangle$ start by $x\in G$. Then there are $a,b\in\mathbb{Z}$ s.t. $x=a\frac{2}{3}+b\frac{5}{7}=\frac{2a}{3}+\frac{5b}{7}=(14a+15b)\frac{1}{21}\in \langle1/21\rangle$. So we have $G\subseteq \langle1/21\rangle$. But we also have $\frac{1}{21}=\frac{5}{7}-\frac{2}{3}$ hence $\langle1/21\rangle\subseteq G$ which means that $G=\langle1/21\rangle$
A: If two integers $a$ and $b$ are coprime, then there are two integers $m$ and $n$ (dependent on $a$ and $b$) such that $ma+nb=1$ (corollary of Bézout's identity). But then, for every integer $k$, $kma+knb=k$: while $k$ runs over $\Bbb Z$, $km$ and $kn$ run over $m\Bbb Z\subseteq \Bbb Z$ and $n\Bbb Z\subseteq \Bbb Z$, respectively. So, $\{xa+yb\mid x\in m\Bbb Z, y\in n\Bbb Z\}=\Bbb Z$, and hence, a fortiori, $\{xa+yb\mid x,y\in \Bbb Z\}=\Bbb Z$.
Now, $14$ and $15$ are coprime, so: $G=\{2x/3+5y/7\mid x,y\in \Bbb Z\}=$ $\{(14x+15y)/21\mid x,y\in \Bbb Z\}=$ $\{z/21\mid z\in \Bbb Z\}=$ $\langle 1/21\rangle$.
