Computing factor groups of infinite groups

Question

Computing factor groups when groups are of finite order is no problem, but when the group is of infinite order I struggle. I have looked at many solutions and really(!) tried to understand them with no luck. In my attempt to understand this, I saw that there are different ways to go about this. Some look at the cosets, and others use the first isomorphism theorem. No matter how hard I try, I am not able to understand any of them. For example $( \mathbb Z \times \mathbb Z \times \mathbb Z_8 )$ $/ <(0, 4, 0)> $ or $(\mathbb Z \times \mathbb Z \times\mathbb Z)$$/<(3, 3, 3)>$ or $(\mathbb Z \times \mathbb Z)$ $/ <(1 , 2)>.$ These are just examples from my book (and I know what the answer is going to be).

My questions are; how does one go about to solve this (in the easiest, beginner friendly way)? I need a dummies step by step explanation, because I dont see the point of the steps taken.

Answer 1

To explain this concept, I'm going to explain $(\Bbb{Z} \times \Bbb{Z} \times \Bbb{Z})/\left<(3, 3, 3)\right>$ because I think it's the easiest to understand.

Let's try to understand this in terms of a few elements. Let's say we have the identity $(0, 0, 0)$. What elements are in the coset $(0, 0, 0)+\left<(3, 3, 3)\right>$? Well, since $(0, 0, 0)$ is just the identity, the coset simply becomes $\left<(3, 3, 3)\right>$, so any element in the normal subgroup is part of the coset that the identity is in.

Now let's look at $(0, 0, 1)$. Is this in the same coset as $(0, 0, 0)$? Well, no, because there's no way $(0, 0, 1)$ can be found by some multiple of $(3, 3, 3)$. Well, then, what elements are in this coset? We can add $(3, 3, 3)$ to it to find $(3, 3, 4)$. We can subtract $(3, 3, 3)$ from it to find $(-3, -3, -2)$. The normal subgroup $\left<(3, 3, 3)\right>$ contains any element in the form of $(3z, 3z, 3z)$ for $z \in \Bbb{Z}$ and we can add that to $(0, 0, 1)$ to find that it shares a coset with all elements in the form of $(3z, 3z, 3z+1)$ for $z \in \Bbb{Z}$.

Now, by a very similar argument, we can understand the coset containing $(0, 0, 2)$ as all of the elements in the form $(3z, 3z, 3z+2)$ and the coset containing $(3, 2, 1)$ as all of the elements in the form $(3z+3, 3z+2, 3z+1)$.

However, sometimes, this won't be the best way to understand the coset. For example, if we use this with $(3, 3, 4)$, we get $(3z+3, 3z+3, 3z+4)$. However, if we subtract $(3, 3, 3)$ from this, we get $(3z, 3z, 3z+1)$ which shows that this is in the same coset as $(0, 0, 1)$, as we found earlier. Our question now is: For which elements can we simplify the coset like this?

Well, in this example, we subtracted $(3, 3, 3)$ because all of the elements were greater than or equal to $3$. Therefore, we can rule out all of those elements. Also, if we had an element like $(-1, -2, -3)$, we would add $(3, 3, 3)$ so we weren't working with negative numbers. Therefore, we can rule out all of those elements.

Thus, if all of the elements can't be negative and all of the elements can't be greater than or equal to $3$, there must be some element that is $0 \leq x < 3$. This element can come from $\Bbb{Z}_3$. However, once we have that element, we can choose the other two elements to come from wherever they want because we have satisfied the conditions above once we have one element from $\Bbb{Z}_3$. Therefore, those other two elements can come from $\Bbb{Z}$. Thus, we have two elements from $\Bbb{Z}$ and one element from $\Bbb{Z}_3$, which results in a factor group of $\Bbb{Z} \times \Bbb{Z} \times \Bbb{Z}_3$.

This kind of exploration with these factor groups is how I like to understand them. I just kind of test certain elements and try to find patterns in where I can simplify my understanding of the cosets. Eventually, you'll learn easier ways to do this where you don't have to take as long and do so much experimentation, but for now, just try to explore the cosets and get a better understanding of them so you can compute the factor group.

Answer 2

Let's look at the first one. What we really want to know is:

"When are two cosets $(a,b,[c]_8) + H$ and $(a',b',[c']_8) + H$ equal for the subgroup $H = \langle(0,4,[0]_8)\rangle$?"

By definition, when $(a,b,[c]_8) - (a',b',[c']_8) \in H$.

So this means:

$a = a'\\b - b' = 4k\\ [c]_8 = [c']_8.$

So, unless the first and the third "coordinates" match, we get different cosets. If these both match, we STILL get different cosets, unless the two "middle" coordinates are equivalent modulo $4$. In other words, to list ALL the cosets, we can:

1. Pick any integer for our first coordinate, any any integer modulo $8$ for our third coordinate, and:

2. Pick any equivalence class modulo $4$ for the middle coordinate.

While it is impossible to list all of these, we can "almost" do it, they are of the form:

$(a,[b]_4,[c]_8) + H$, which should strongly suggest the factor group is (isomorphic to):

$\Bbb Z \times \Bbb Z_4 \times \Bbb Z_8$

I do not know if you know the Fundamental Homomorphism Theorem, but if you do, we can reason this way:

Define $\phi:\Bbb Z \times \Bbb Z \times \Bbb Z_8 \to \Bbb Z \times \Bbb Z_4 \times \Bbb Z_8$ by:

$\phi(a,b,[c]_8) = (a,[b]_4,[c]_8)$ (basically, we keep everything the same but reduce the middle coordinate mod $4$).

I won't prove this is a homomorphism (I urge you to do so, though, it's not that hard). Now suppose $x \in \langle (0,4,[0]_8)\rangle$, so that: $x = (0,4k,[0]_8)$ for some integer $k$.

Then $\phi(x) = \phi(0,4k,[0]_8) = (0,[4k]_4,[0]_8) = (0,[0]_4,[0]_8)$, the identity of $\Bbb Z \times \Bbb Z_4 \times \Bbb Z_8$. this shows that $H \subseteq \text{ker }\phi$.

On the other hand, if $\phi(a,b,[c]_8) = (0,[0]_4,[0]_8)$ (that is $(a,b,[c]_8) \in \text{ker }\phi$), by the definition of $\phi$ we have:

$a = 0\\ [b]_4 = [0]_4\\ [c]_8 = [0]_8.$

The middle equation tells us $b = 4k$ (since $[b]_4 = [0]_4$), so we have:

$(a,b,[c]_8) = (0,4k,[0]_8) = k(0,4,[0]_8) \in H$.

Thus $\text{ker }\phi \subseteq H$, and subsequently, $\text{ker }\phi = H$.

Evidently, $\phi$ is onto (surjective-why?), so the Isomorphism Theorem tells us:

$(\Bbb Z \times \Bbb Z \times \Bbb Z_8)/H \cong \Bbb Z \times \Bbb Z_4 \times \Bbb Z_8$