Using the first isomorphism theorem I'm having trouble using the first isomorphism theorem to show that
$$ \Bbb Z \times \Bbb Z \times \Bbb Z / \langle (4,4,4) \rangle$$
is isomorphic to 
$$ \Bbb Z \times \Bbb Z \times \Bbb Z_4$$
How would I apply the theorem to show the isomorphism in this case?
 A: One way to do this is is the following.
Let's construct a  surjective homomorphism $T: \mathbb Z \times \mathbb Z \times \mathbb Z \to \mathbb Z \times \mathbb Z \times \mathbb Z/4\mathbb Z$ so that $(4, 4, 4)$ is sent to $(0, 0, 4) \cong (0, 0, 0)$ in $\mathbb Z \times \mathbb Z \times \mathbb Z/4\mathbb Z$, i.e. $T(4, 4, 4) = (0, 0, 4)$. Now the simplest way to extend this as a linear map to the whole domain is to send $(1, 0, 0)$ to itself and $(0, 1, 0)$ to itself. Now you can check that this implies $T(1, 0, 0) = (-1, -1, 1)$ and so you can easily compute images of $T$. The kernel is $\langle (4, 4, 4) \rangle$. Now apply the first isomorphism theorem.
If you like, you can also treat it like a linear algebra problem and write the matrix of $T$ in the standard basis,
$$
[T] =
\begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{pmatrix}.
$$
Now this would usually have trivial kernel since it has full rank but since vectors mapped to multiples of $(0, 0, 4)$ are also in the kernel, the kernel is $\langle (4, 4, 4) \rangle$ as we had constructed.
