Show that the following group $H$ is a cyclic group Let $a,b$ be fixed positive integers and $H=\{ax+by:x,y\in \mathbb Z \}$. Show that $H$ is a cyclic group with $\gcd(a,b)$ as a generator.
Proceed
Let $d=\gcd(a,b)$.  What can I say after this?    
Added
 A group $G $ is said to be cyclic if   there exists an element $a$ in $G$ such that 
$$G=\{ a^n: n\in \mathbb Z\}$$
 A: *

*If $d$ is any common divisor of $a$ and $b$, then $a = kd$ and $b = \ell d$ for some integers $k$ and $\ell$.
So for any element $ax + by \in H$, $$ax + by = (kd)x + (\ell d)y = (kx+\ell y)d$$
which shows that every element in $H$ is a multiple of $d$.


*If $d$ is the greatest common divisor of $a$ and $b$, then (Bézout's identity) there exist integers $x$ and $y$ such that $ax + by = d$. Multiply this equation by any integer $n$; we have that $a(nx) + b(ny) = nd$, which shows that every multiple of $d$ is in $H$.


*If every element in $H$ is a multiple of $d$, and every multiple of $d$ is in $H$, it follows that $H$ is exactly the set of all multiples of $d$:
$$H \equiv \{nd : n\in \mathbb{Z}\}$$
Q.E.D.
A: We need to establish that $H$ is a group.
And that every element in the group, can be expressed as $nd$ 
$d = \gcd(a,b)$ implies that there exist some $x,y \in \mathbb Z$ such that $ax + by = d$
Since $ax + by = d \implies d \in H$
and
$a(2x) + b(2y) = 2d \implies 2d \in H$
$nd \in H \implies nd + d = (n+1)d \in H$
$0$ is the identity 
$-d + d = 0 \implies -d$ is the inverse of $d$ and every element $nd$ has an inverse element$-nd.$
We have established that $H$ is a group.
Now we need to show that there is no element in $H$ that cannot be expressed as $nd$
By the definition of $\gcd(a,b)$ there is no $x,y$ such that $0<ax + by < d$
Suppose there is some element in $h\in H$ that not equal to $nd$
and $x', y'\in \mathbb Z$ such that $ax' + by' = h$
Then for some $n, nd<h<(n+1)d$
and $h-nd < d$
then $a(x'-nx) + b(y'-ny)<d$
Contradicting that there is no $x,y$ such that $0<ax + by < d$
