I want to compute following things $(\mathbb{Z} \oplus \mathbb{Z} ) / \langle(3,6),(5,5)\rangle$
First of all, I know for one generator case $(\mathbb{Z} \oplus \mathbb{Z} ) / \langle(m,n)\rangle \cong \mathbb{Z}$
If $gcd(m,n)=1$ and for $gcd(m,n)=d$, it reduces $\mathbb{Z}\oplus \mathbb{Z}_d$
What I want to do next is generalized this case to two generators. Is there any ideas for computing this kinds of things?
On
SNF on the generators matrix: $$\begin{pmatrix}3&6\\5&5\end{pmatrix}\stackrel{L_2-2L_1}\longrightarrow\begin{pmatrix}3&0\\5&\!-5\end{pmatrix}\stackrel{R_2-\frac53R_1}\longrightarrow\begin{pmatrix}3&0\\0&\!-5\end{pmatrix}\longrightarrow\begin{pmatrix}3&0\\0&5\end{pmatrix}$$
and you get a(n abelian) group of order
$$\;\det\begin{pmatrix}3&0\\0&5\end{pmatrix}=15\;,\;\;\text{and in fact}\;\;\Bbb Z\oplus\Bbb Z/\langle(3,6),\,(5,5)\rangle\cong\Bbb Z_3\oplus\Bbb Z_5$$
Note that $$\begin{aligned} \langle (3,6),(5,5)\rangle &=\langle (5,5) -2(3,6), (3,6) \rangle = \langle (1,7),(3,6) \rangle \\ &= \langle (1,7), (3,6)-3(1,7) \rangle = \langle (1,7), (0,15) \rangle \end{aligned}$$
Therefore, $$(\mathbb{Z}\oplus \mathbb{Z})/\langle (3,6),(5,5)\rangle \cong (\mathbb{Z}\oplus \mathbb{Z}_{15})/\langle (1,\overline{7})\rangle := G$$
I put over-line on the seven to emphasis it is computed modulo 15. Now lets interpret the last group (or ring). In $G$, any $(a,b)\in \mathbb{Z}\times \mathbb{Z}_{15}$ equals $(0,c)\in \mathbb{Z}\times \mathbb{Z}_{15}$ for some $c$. Note that $$a = (1,\overline{0}) = (0,\overline{-7})$$
You will convince yourself that $G\cong C_{15}$ after computing $2a, 3a, ..., 14a$.
If you know something about $\mathbb{Z}$-module, then the abelian group given has presentation matrix $\begin{pmatrix} 3 & 5 \\ 6 & 5 \end{pmatrix}$, which can be brought to $\begin{pmatrix} 5 & 0 \\ 0 & 3 \end{pmatrix}$ by some elementary column and row operation (in $\mathbb{Z}$), so $G\cong C_3\times C_5$.