I am following a solution to classify $G/H$, $G=\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$, $H=\langle (1,1,2)\rangle$, according to the theorem of finitely generated abelian groups. This is my first course in abstract algebra and I am at the chapter of factor groups.
It claims that all the cosets of $G/H$ are $(m,n,0)+H$ and $(m,n,1)+H$. The way I showed they are all the cosets is:
If $c$ is even: $(a, b, c) - c/2(1, 1, 2) = (a-c/2, b-c/2, 0)$ of the form $(m,n,0)$.
If $c$ is odd: $(a, b, c) - (c-1)/2(1,1,2)=(a-(c-1)/2,b-(c-1)/2,1)$ of the form $(m,n,1)$.
where $(1,1,2)\in H$.
- Is this correct? Why do I show they are all cosets in this way? How do I come up with $(m,n,0)$ and $(m,n,1)$?
To show these two cosets are different, I computed: $(m,n,1)-(m,n,0)=(0,0,1)$ which is not in $H$, knowing that $ab^{-1}\in H$ if and only if $a\in bH$.
The solution finally says that $\mathbb{Z}^{2}$ is included in $G/H$, but I don't understand why this is the case. The elements of these two sets are different.
EDIT: I don't know why my question was marked as duplicate to... my other question? These are two very different questions with two different methods. If these two are the same then no other $\mathbb{Z}^3/H$ question will be allowed in this site?
First of all, I recommend that you forget about cosets. Quotient groups are not really about cosets. When you encounter a quotient group, the following definition* is very useful.
Very much related to this definition is the fundamental theorem on homomorphisms, or equivalently, the universal property of quotient groups. Cosets provide a possible construction of quotient groups (there are others as well), but in practice the mentioned definition is much more useful. Computer scientists would call cosets an implementation detail.
*Strictly speaking, this does not specify $G/N$ in a unique way (with the traditional meaning of that word), but any two quotient groups, defined that way, will be canonically isomorphic to each other. And that's enough (for structural mathematics). This also justifies taking this, what many just call a characterization, as a definition.
So when you want to determine a quotient group $G/N$, the only thing you gotta do is to find a group $Q$ with a surjective homomorphism $G \to Q$ with kernel $N$. Then $Q$ will be one model of $G/N$.
Let us consider the example $\mathbb{Z}^3 / (0 \times 0 \times \mathbb{Z})$. We need to find a surjective homomorphism $\mathbb{Z}^3 \to {???}$ with kernel $0 \times 0 \times \mathbb{Z}$. Well, the homomorphism $\mathbb{Z}^3 \to \mathbb{Z}^2$, $(a,b,c) \mapsto (a,b)$ is clearly surjective and has kernel $0 \times 0 \times \mathbb{Z}$. Hence, the quotient group is realized by $\mathbb{Z}^2$.
Next, we have the following basic result (which follows immediately from the fundamental theorem on homomorphisms): If $f : G \to G'$ is an isomorphism of groups, $H \subseteq G$ is a normal subgroup, $H' := f(H)$, then there is a unique isomorphism $\overline{f} : G/H \to G'/H'$ which is characterized by $\overline{f}([g]) = [f(g)]$, where I denote the canonical projection $G \to G/H$ by $g \mapsto [g]$.
There is an isomorphism $f : \mathbb{Z}^3 \to \mathbb{Z}^3$ that maps $\langle (1,1,2) \rangle$ onto $0 \times 0 \times \mathbb{Z}$, namely $(a,b,c) \mapsto (b-a,c-2a,a)$. It is clearly a homomorphism with that property, and it is an isomorphism since you can define an inverse: $(u,v,w) \mapsto (w,u+w,v+2w)$.
The basic result hence implies $$\mathbb{Z}^3 / \langle (1,1,2) \rangle \cong \mathbb{Z}^3 / (0 \times 0 \times \mathbb{Z}) \cong \mathbb{Z}^2.$$
By the way, we can generalize the result: If $(a_1,\dotsc,a_n) \in \mathbb{Z}^n$ is arbitrary, then $\mathbb{Z}^n / \langle (a_1,\dotsc,a_n) \rangle \cong \mathbb{Z}^{n-1} ~ \oplus ~ \mathbb{Z}/\mathrm{gcd}(a_1,\dotsc,a_n)$.