Let a, b be fixed positive integers and $H=\{ax+by|x,y\in \Bbb Z\}.$ Show that $H$ is a cyclic group with $\gcd(a,b)$ as a generator. 
Let $a$, $b$ be fixed positive integers and $$H=\{ax+by|x,y\in \Bbb Z\}.$$ Show that $H$ is a cyclic group with $\gcd(a,b)$ as a generator.

Approach:
Let $d=\gcd(a,b)$
then $d|a, ~d|b$
i.e $d\alpha =a, ~d\beta =b$ for some integers $\alpha, \beta$
Let $ax+by\in H$ then
$ax+by=d(x\alpha + y\beta)\in dZ$ i.e $$H\subset d\Bbb Z ~~~~~~~~~~~~~(1)$$
We now show that $d \Bbb Z \subset H$
By Euclidean algorithm, there exist $u, v$ such that
$ua+vb=\gcd(a,b)=d$
Please correct my steps and help me to complete the remaining part of solution. I am unable to proceed further. 
EDIT: Question in clear form:

I would like to show that $$d\mathbb Z =H.$$ I have shown $$H\subset d\Bbb Z$$ and I am unable to show  $$d \Bbb Z \subset H$$ Please help me to show the desired part ($d \Bbb Z \subset H$).

 A: First show that $H$ is a subgroup of $\mathbb Z$ which is easy to show. Then use the fact (?) that any subgroup of $\mathbb Z$ is of the form $n\mathbb Z$ for some integer $n$. Hence $H$ is a cyclic group, of the form $d\mathbb Z$ for some $d\in\mathbb Z$.
We claim $d=gcd(a,b)$.
Note that $a=a.1+b.0\in H$ and $b=a.0+b.1\in H$ so $a,b\in d\mathbb Z$ implying $d|a,d|b$ implying $d$ is a common divisor of $a$ and $b$.
Now $d\in d\mathbb Z$ so there exist integers $x,y$ so that $d=ax+by$. Let $k$ be a common divisor of $a$ and $b$ then this equation shows that $k$ divides the RHS, so $k$ divides LHS i.e. $k|d$.
These conclude that $d=gcd(a,b)$.
A: Let $d=\gcd(a,b)$. Then $a=d\cdot r$ and $b= d\cdot s$ for some integers $r, s$.
Now $\gcd(r,s)=1$. Then $\exists\ r\cdot u+s\cdot v=1$ for some integers $u, v$.
So for any integer $z,\ z=r\cdot u\cdot z+s\cdot v\cdot z$.
Hence $d\cdot z= d\cdot r\cdot u\cdot z+d\cdot s\cdot v\cdot z=a\cdot u\cdot z+b\cdot v\cdot z=a\cdot x+b\cdot y$, where $x=u\cdot z$ and $y=v\cdot z$ are two integers.
Hence $d\Bbb Z \subset H$.
