Calculate factor group $(\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z})/\langle(1,1,1)\rangle$

Question

My instructor's said at the lecture that you basically set one of the dimensions to $0$ and hence you get $\mathbb{Z} \times \mathbb{Z}$. Again, is there a better way to think about this problem and approach it? I am only a novice abstract algebra student and his instructions are way too chaotic here.

Answer 1

Your instructor's intuition is correct, but we can formalize this by writing down an explicit isomorphism $\tilde{\Phi}: \mathbb{Z}^3 / \langle(1, 1, 1)\rangle \to \mathbb{Z}^2$.

First, given an element $(a, b, c) \in \mathbb{Z}^3$, it is equivalent to exactly one element of the form $(x, y, 0)$, namely $$(a, b, c) + (-c) \cdot (1, 1, 1) = (a - c, b - c, 0),$$ and so each coset contains exactly one element of that form.

Now, the set of elements of the form $(x, y, 0)$ obviously form a group under addition isomorphic to $\mathbb{Z}^2$, via the map $(x, y, 0) \mapsto (x, y)$, which suggests we consider the homomorphism $\Phi: \mathbb{Z}^3 \to \mathbb{Z}^2$ defined by $$\Phi: (a, b, c) \mapsto (a - c, b - c).$$ By construction, $\Phi$ is constant on the cosets, so it factors to a map $\tilde{\Phi}: \mathbb{Z}^3 / \langle(1, 1, 1)\rangle \to \mathbb{Z}^2$. This map is surjective because $\Phi$ is, and it is injective because of our observation that each coset contains exactly one element of the form $(x, y, 0)$, so it is an isomorphism. (Of course, we can also check these latter two claims directly.)