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I'm studying quotient groups (by myself) and having a hard time with them.

I will reference

This question is more of a followup on the first link.

I'm interested in infinite quotient groups. The exercises I am doing are $\mathbb{Z}^3/\left<(1,1,1)\right>$ and $\mathbb{Z}^3/\left<(3,0,0)\right>$ but I feel like I'm missing some concepts to do them. I have to write the quotient groups and classify them with the fundamental theorem of finitely generated abelian groups. I look at groups of the form $\mathbb{Z}^3/\left<(x,y,z)\right>$.

Starting with the definition, $\mathbb{Z}^3/H$ is the set of all elements $gH$ with $g\in \mathbb{Z}^3$: $\{(0,0,0)+H, (0,0,1)+H, ..., (i,j,k)+H,...\}$. When $H=\left<(1,1,1)\right>=\left<(x,x,x)\right>$, $\mathbb{Z}^3/H=\{(a,b,c)+\left<(a,a,a)\right>\}$.

First: why do I need to find generators for $\mathbb{Z}^3/H$? Is this because the group is cyclic, to make it easier to classify? The answers linked above start by: $\mathbb{Z}^3$ has basis $(1,0,0),(0,1,0),(0,0,1)$ but also $(1,1,1),(0,1,0),(0,0,1)$ generates $\mathbb{Z}^3$. Why is this important?

I understand that the matrix with those three vectors can be transformed into the identity matrix (which again is the canonical basis), but why? Is it always this obvious to transform such a matrix? If I had $\mathbb{Z}^3/\left<(3,0,0)\right>$, how would I go from $(3,0,0)$ to a matrix "equivalent" to the identity matrix? I read about the Smith normal form; is there a way without it?

Now, I have generators for $\mathbb{Z}^3$ and for $\left<(1,1,1)\right>$ (that element). How do I find the generators for $\mathbb{Z}^3/H$?

Shaun
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hirdajarzu
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    See https://en.wikipedia.org/wiki/Smith_normal_form – lhf Nov 23 '23 at 12:28
  • @lhf thanks but I mention I want to do these exercises without it – hirdajarzu Nov 23 '23 at 12:42
  • Why should you want to do these exercises without the correct tool? – ancient mathematician Nov 23 '23 at 14:57
  • @ancientmathematician because this is not the correct tool for me. The book has not yet covered it, so I want to do the exercises without it. Plus asking for another or simpler approach is fine isn't it? – hirdajarzu Nov 23 '23 at 15:20
  • Yes, textbooks do sometimes make silly demands. Perhaps there is some clever way of doing these special cases - but the Smith Normal Form algorithm will do all cases without any effort. – ancient mathematician Nov 23 '23 at 15:57
  • @ancientmathematician great! I want to do this with a more elementary approach. This will help me develop a better understanding of the topic. – hirdajarzu Nov 23 '23 at 16:02

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I will try to use only elementary facts about groups and homomorphisms. At all stages where I assert that a map is a homomorphism, is onto, has some kernel that needs to be checked.

So let's fix notation $e_1:=(1,0,0),e_2:=(0,1,0), e_3:=(0,0,1)$.

$\mathbb{Z}^3/\langle (1,1,1)\rangle$:

That is we are looking at $\langle e_1, e_2, e_3\rangle/\langle (e_1+e_2+e_3)\rangle$.

In this case let $f_1:=(1,1,1),f_2:=(0,1,0), f_3:=(0,0,1)$. Then note that $e_1=f_1-f_2-f_3, e_2=f_3, e_3=f_3$. That means that $\langle f_1,f_2,f_3\rangle=\langle e_1,e_2,e_3\rangle$ (and not just some subgroup).

So in fact we are looking at the quotient $\langle f_1, f_2, f_3\rangle/\langle f_1\rangle$. If we look at the homomorphism $\sigma: \langle f_1, f_2, f_3\rangle\to \langle f_2, f_3\rangle$ given by $\sigma: a_1 f_1+a_2 f_2 +a_3 f_3\mapsto a_2 f_2 +a_3 f_3$ we see that it is onto and has kernel $\langle f_1\rangle$. So by the Isomorphism Theorem we have that $$ \langle f_1, f_2, f_3\rangle/\langle f_1\rangle\simeq \langle f_2,f_3\rangle\simeq \mathbb{Z}^2. $$

$\mathbb{Z}^3/\langle (3,0,0)\rangle$:

Note that $\tau: \langle e_1,e_2,e_3\rangle\to \langle e_1\rangle/\langle 3e_1\rangle \times \langle e_2,e_3\rangle$ by $\tau:a_1 e_1+a_2e_2+a_3 e_3\mapsto (a_1 e_1+\langle 3 e_1\rangle, a_2 e_2 +a_3 e_3)$ is a homomorphism, onto, and has kernel $\langle 3e_1\rangle$.

Hence $$ \langle e_1,e_2,e_3\rangle/\langle 3e_1\rangle \simeq \langle e_1\rangle/\langle 3e_1\rangle \times \langle e_2,e_3\rangle \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}^2. $$

ancient mathematician
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  • How do you find that $\tau$ maps to an element of the form $(a, b)$ and not, say, $m\in \mathbb{Z}$ or anything else? I know that if you do it like that, it works; but how do you get the idea? – hirdajarzu Dec 01 '23 at 07:51
  • I got the idea because the general Smith Normal Form Algorithm says something like: you can find new generators of $G$ call them $f_1,f_2,\dots, f_m$ such that the generators of $N$ are multiples of these, and then it's easy to calculate $G/N$. Seriously if you are interested in the topic learn the right way to do it and don't worry about these special cases. – ancient mathematician Dec 01 '23 at 12:28