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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?

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    $\begingroup$ You should consider the map $\varphi\colon \Bbb Z \times \Bbb Z \times \Bbb Z \to \Bbb Z \times \Bbb Z \times \Bbb Z_4 \rangle$ given by $(a,b,c)\mapsto (a-c,b-c, \bar{c})$; show it is homomorphism, and calculate $\ker\varphi$ and $\text{im}\varphi$. $\endgroup$ – ancientmathematician Mar 13 '17 at 7:28
  • $\begingroup$ Yeah, I just noticed that such $\varphi$ was wrong, so I deleted my comment just before your comment appeared. $\endgroup$ – EternalBlood Mar 13 '17 at 7:42
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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.

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