Calculate the quotient groups and classify $\mathbb{Z^3}/(1, 1, 1)$ - Fraleigh p. 151 15.8

Question

I tried to start like this question. I fill in the sizes first. $\mathbb{Z}$ is an infinite group hence $\mathbb{Z^3}$ is too. $\left|\dfrac{\mathbb{Z^3}}{<(1, 1, 1)>}\right| = \dfrac{|\mathbb{Z^3}|}{|<(1, 1, 1)> \in \mathbb{Z^3} |} = \frac{\infty}{\infty}$. $\color{darkred}{ \text{ (1.) What did I bungle here? } }$

(1.) Now I tried http://www.physicsforums.com/showthread.php?t=355789 but I'm confounded. Can someone please flesh out all the steps? Can this be done without fundamental theorem of finitely generatedabelian groups ?

(2.) What's the intuition, by means of the last paragraph here?

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Questions on Christoph Pegel's answer: I deleted my original try.

I'm not sure why you want to calculate the order of the quotient first, but for infinite groups you can not just divide the orders. For example $\left|\mathbb Z\big/2\mathbb Z\right|=2$ while $\left|\mathbb Z\right|=\left|2\mathbb Z\right|=\infty$.

Now let's look at $\mathbb Z^3\big/ H$ where $H=\langle\left(1,1,1\right)\rangle$. The idea is that you can always make the first coordinate $0$ by adding or substracting $(1,1,1)$ multiple times. To be precise: $$ (x,y,z) - x\cdot(1,1,1) = (x,y,z) - (x,x,x) = (0, y-x, z-x). $$

Update (3.) I understand this proves every coset has a representative of the form $(0,y,z).$
Hence you're authorized to drop the generator $(1, 0, 0) \in \mathbb{Z}_3$.

But before you proved this, how do you envisage and envision this hinge/'key observation'? This feels magical. What induced you to make the first coordinate $0$ by subtracting $(1, 1, 1)$ $x$ times?

So for all cosets $[(x,y,z)]\in \mathbb Z^3\big/ H$ you have $[(x,y,z)] = [(0, y-x, z-x)]$ in the quotient. Every coset has a representative with $0$ in the first coordinate.

Now can a coset have multiple representatives of this form? Assume $$ (0,y,z)H=(0,y',z')H \iff \quad (0,y,z) - (0,y',z') = (\color{dodgerblue}0, \color{green}{y-y'}, \color{red}{z-z'}) \qquad \qquad \in H = \langle(1,1,1)\rangle. $$ All elements in H = $< (1, 1, 1) > = k(1, 1, 1) = (\color{dodgerblue}k,\color{green}k,\color{red}k) $ for all $k \in \mathbb{Z}$.
Equate component-wise to induce: $\color{dodgerblue}{k = 0}, \color{green}{y - y' = k}, \color{red}{z -z' = k}.$
Substitute $\color{dodgerblue}{k = 0}$ to induce: $\implies \color{green}{y - y' = 0}, \color{red}{z -z' = 0}$.
Thus $y=y'$ and $z=z'$ and the representatives are equal.

(4.) What do your square brackets mean? Are they my curly ones $\{$?

To summarize what we know so far:
Every coset in $\mathbb Z^3\big/ H$ has exactly one representative of the form $(0,y,z)$.

This gives a bijective map $ \begin{align*} \Phi : \mathbb Z^2 &\longrightarrow \mathbb Z^3\big/ H, \\ (y,z) &\longmapsto [(0,y,z)]. \end{align*}$

(5.) Can you please unfold where this bijective function magically sprang from?

Now you can easily show this map is a homomorphism, so indeed
(My original post had the answer) $\mathbb Z^3\big/ H \cong \mathbb Z^2$.

Answer 1

I'm not sure why you want to calculate the order of the quotient first, but for infinite groups you can not just divide the orders. For example $\left|\mathbb Z\big/2\mathbb Z\right|=2$ while $\left|\mathbb Z\right|=\left|2\mathbb Z\right|=\infty$.

Now let's look at $\mathbb Z^3\big/ H$ where $H=\langle\left(1,1,1\right)\rangle$. The idea is that you can always make the first coordinate $0$ by adding or substracting $(1,1,1)$ multiple times. To be precise: $$ (x,y,z) - x\cdot(1,1,1) = (x,y,z) - (x,x,x) = (0, y-x, z-x). $$ So for all cosets $[(x,y,z)]\in \mathbb Z^3\big/ H$ you have $[(x,y,z)] = [(0, y-x, z-x)]$ in the quotient.
Every coset has a representative with $0$ in the first coordinate.

Now can a coset have multiple representatives of this form? Assume $[(0,y,z)]=[(0,y',z')]$, then $$ (0,y,z) - (0,y',z') = (0, y-y', z-z') \in H = \langle(1,1,1)\rangle. $$ All elements in $H$ are of the form $(k,k,k)$ for some $k\in \mathbb Z$, so from $(0, y-y', z-z') \in H$ it follows that $y-y'=0$ and $z-z'=0$ since the first coordinate is $0$. Thus $y=y'$ and $z=z'$ and the representatives are equal.

To summarize what we know so far:
Every coset in $\mathbb Z^3\big/ H$ has exactly one representative of the form $(0,y,z)$.

This gives a bijective map
\begin{align*} \Phi : \mathbb Z^2 &\longrightarrow \mathbb Z^3\big/ H, \\ (y,z) &\longmapsto [(0,y,z)]. \end{align*}

Now you can easily show this map is a homomorphism, so indeed $\mathbb Z^3\big/ H \cong \mathbb Z^2$.

Regarding your questions:

(3.) Can you please flesh out this idea? Why do we want the first coordinate 0? This looks like the hinge. I'm perplexed where this loomed from.

This is indeed the key observation. While $\mathbb Z^3$ is generated by the 3 generators $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, we can generate $\mathbb Z^3\big/H$ by the 2 generators $[(0,1,0)]$ and $[(0,0,1)]$, since (as I showed) every coset has a representative of the form $(0,y,z)$. So we can drop a generator, which suggests that our quotient is isomorphic to $\mathbb Z^2$ (or a quotient of that).

(4.) What do your square brackets mean? Are they my curly ones {?

No, the curly ones are always sets. The square brackets denote cosets: $$ [(x,y,z)] := (x,y,z) + H. $$

(5.) Can you please unfold where this bijective function magically sprang from?

It's the whole idea of the proof. We showed that we only need the two generators $[(0,1,0)], [(0,0,1)]$. So let's take $\mathbb Z^2$ and map it's two generators $(1,0)$, $(0,1)$ to the two of $\mathbb Z^3\big/H$ and we get an isomorphism!

Indeed if we have any abelian group $G$ generated by 2 elements $G=\langle a,b\rangle$ we get an epimorphism by \begin{align} \Phi: \mathbb Z^2 &\longrightarrow G,\\ (x,y) &\longmapsto x\cdot a + b\cdot y. \end{align} Try to prove this! Then, by the homomorphism theorem, $\mathbb Z^2/\ker\Phi\cong G$. In our case $\Phi$ is injective so $\ker\Phi=\{(0,0)\}$ and thus $\mathbb Z^2\cong G$.

Update (3.) I understand this proves every coset has a representative of the form (0,y,z). Hence you're authorized to drop the generator (1,0,0)∈ℤ3.

But before you proved this, how do you envisage and envision this hinge/'key observation'? This feels magical. What induced you to make the first coordinate 0 by subtracting (1,1,1) x times?

It's an idea, thats very different from magic. And as you see, this idea turns out just fine. Maybe start with $\mathbb Z^3\big/\langle(1,0,0)\rangle$ which gives you $[(x,y,z)]=[(0,y,z)]$ immediatly, since we just modded out a generator. For $\mathbb Z^3\big/ H$ we are in a similar situation: We can generate $\mathbb Z^3$ by $(1,1,1)$, $(0,1,0)$ and $(0,0,1)$ (check this!), modding out one of these leaves us with the other two.

(2.) What's the intuition, by means of the last paragraph here?

When you look at the quotient $\mathbb Z\times\mathbb Z^6\big/\langle(1,2)\rangle$ you get the relation $[(1,2)]=[(0,0)]$ which you could also write as $[(1,0)]=[(0,-2)]=[(0,4)]$. Since $(0,4)$ has order $3$, the order of $[(1,0)]$ in the quotient must divide $3$ as well.