I asked an earlier question (Finding elements of infinite order in quotient groups) and I realize I am having trouble understanding the quotient groups in the questions posted. Hence I am posting a separate more basic question related to it.
The quotient group $\mathbb{Z}/\langle n \rangle$, the cosets are of the form $\mathbb{Z}/\langle n \rangle=\{0 + \langle n \rangle, 1 + \langle n \rangle, \ldots ,(n-1) + \langle n \rangle\}$, and there are $n$ of them.
In the case of $(\mathbb{Z}\times \mathbb{Z})/\langle (a,b) \rangle$, if $a=b$, then there should be $a^2$ number of cosets? They are of the form $(\mathbb{Z}\times \mathbb{Z})/\langle (a,a) \rangle=\{(0,0)+ (\langle a \rangle \times \langle a \rangle), \ldots ,(a-1,a-1)+ (\langle a \rangle \times \langle a \rangle)\}$
if $a\neq b$ and $gcd(a,b)=1$, then there should be $ab$ number of cosets in $(\mathbb{Z}\times \mathbb{Z})/\langle (a,b) \rangle$ The cosets are of the form $(\mathbb{Z}\times \mathbb{Z})/\langle (a,b) \rangle=\{(0,0)+ (\langle a \rangle \times \langle b \rangle), \ldots ,(a-1,b-1)+ (\langle a \rangle \times \langle b \rangle)\}$
Finally if $a\neq b$ and $gcd(a,b)\neq 1$, then can I reduce $\frac{a}{b}=\frac{a'}{b'}$, where $gcd(a',b')= 1$. So the number of cosets in the quotient group $(\mathbb{Z}\times \mathbb{Z})/\langle (a,b) \rangle$ is $a'b'$ and they are of the form $(\mathbb{Z}\times \mathbb{Z})/\langle (a,b) \rangle=\{(0,0)+ (\langle a' \rangle \times \langle b' \rangle), \ldots ,(a'-1,b'-1)+ (\langle a' \rangle \times \langle b' \rangle)\}$
Thank you in advance.