Determination of isomorphism between finite abelian groups Let $G_1=\dfrac{\mathbb{Z}_{5^{20}} \times \mathbb{Z}_{5^{16}}  \times \mathbb{Z}_{5^{14}}}{\langle (2, 35, 3)  \rangle}$. The SNF of it is $G_2:=\mathbb{Z}_{5^{16}} \times \mathbb{Z}_{5^{14}}$.
I was curious to derive an isomorphism $f:G_1\rightarrow G_2$ explicitly. Here is what I have done so far:
Now I found $\dfrac{\mathbb{Z}_{5^{20}} \times \mathbb{Z}_{5^{16}}}{\langle (2, 35)  \rangle}\simeq \mathbb{Z}_{5^{16}}$. Let $f_1$ be one isomorphism. Also I found
$\dfrac{\mathbb{Z}_{5^{20}} \times \mathbb{Z}_{5^{14}}}{\langle (2, 3)  \rangle}\simeq \mathbb{Z}_{5^{14}}$, so let $f_2$ be an isomorphism. From here I was trying to construct $f$ using $f_1, f_2$ but could not make it.
Is there any other way to find $f$? Or do I need to follow above way?
Please help.
 A: One way to do this is to find a homomorphism:
$$f:\mathbb Z_{5^{20}}\times\mathbb Z_{5^{16}}\times\mathbb Z_{5^{14}}\to \mathbb Z_{5^{16}}\times\mathbb Z_{5^{14}}$$ which is onto and has kernel $(2,35,3).$
Let:
$$f(a,b,c)=(b-35qa,c-3qa).$$
Then $f(a,b,c)=0$ iff $b=35qa, c=3qa$ iff $(a,b,c)=a(1,35q,3q).$
So the kernel of $f$ is $\langle(1,35q,3q)\rangle=\langle(2,35,3)\rangle.$
We get $f$ is onto, since $f(0,b,c)=(b,c).$
So $$\phi((a,b,c)+\langle(2,35,3)\rangle)=(b-35qa,c-3qa)$$ is your isomorphism.

Technically, I am using implicitly the homomorphisms $\alpha_1:\mathbb Z_{5^{20}}\to\mathbb Z_{5^{16}}$ and $\alpha_2:\mathbb Z_{5^{20}}\to\mathbb Z_{5^{14}},$ both of which send $1$ to $1.$
If you want to be more precise:
$$\phi((a,b,c)+\langle(2,35,3)\rangle)=(b-35\alpha_1(qa),c-3\alpha_2(qa))$$

In general, if $m_3, m_2\mid m_1$ and $\gcd(u,m_1)=1$ we can show that:
$$\dfrac{\mathbb Z_{m_1}\times\mathbb Z_{m_2}\times\mathbb Z_{m_3}}{\langle(u,v,w)\rangle}\cong \mathbb Z_{m_2}\times\mathbb Z_{m_3}$$
By finding solution $uq+m_1r=1$ and then sending:
$$(a,b,c)\to (b-vqa,c-wqa)$$
And this, of course, extends to more rings: $m_2,\dots,m_{k}\mid m_1$ and $\gcd(u_1,m_1)=1,$ then:
$$\dfrac{\prod_{i=1}^k \mathbb Z_{m_i}}{\langle(u_1,\dots,u_k)\rangle}\cong \prod_{i=2}^k \mathbb Z_{m_i}$$

Even more generally, if $A_1=\mathbb Z_{m}$ and $A_2$ another abelian groups, and $\gcd(u,m)=1$ and $v\in A_2$ such that $mv=0.$ Then we get:
$$\dfrac{A_1\times A_2}{\langle (u,v)\rangle}\cong A_2$$
Since $mv=0,$ there is a homomorphism $\rho:\mathbb Z_{m}\to A_2$ with $\rho(1)=v.$
Also, $uq=1$ for some $q.$
Then we can show $\langle(u,v)\rangle = \langle(1,qv)\rangle,$ and we define $$f(a,b)=b-\rho(aq).$$
$(a,b)$ is in the kernel when $b=\rho(aq)=aqv$ and thus $(a,b)=a(1,qv).$
In your original question, $\rho(a)=(35a,3a).$
