I'm studying quotient groups (by myself) and having a hard time with them.
I will reference
- Calculate the quotient groups and classify $\mathbb{Z^3}/(1, 1, 1)$ - Fraleigh p. 151 15.8
- Classification of $\mathbb{Z}\times \mathbb{Z} \times \mathbb{Z} / \langle (n,n,n) \rangle$
- Classification of finite abelian groups
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 $\mathbb{Z}^3/H$?
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 $$ \langle e_1,e_2,e_3\rangle/\langle 3e_1\rangle \simeq \langle e_1\rangle/\langle 3e_1\rangle \times \langle e_2,e_3\rangle \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}^2. $$