Exact Sequences - Diagrams that 'commute' - Example

[Math Amateur]

I am reading Dummit and Foote Section 10.5 on Exact Sequences.

I am trying to understand Example 1 as given at the bottom of page 381 and continued at the top of page 382 - please see attachment for the diagram and explanantion of the example.

The example, as you can no doubt see, requires an understanding of the nature of the quotient module $$(\mathbb{Z} / m \mathbb{Z} ) / (n \mathbb{Z} / m \mathbb{Z} )$$

To make this quotient more tangible, in this example take m = 6, n = 3 so k = 2.

Then we are trying to understand the nature of the quotient module $$(\mathbb{Z} / 6 \mathbb{Z} ) / (3 \mathbb{Z} / 6 \mathbb{Z} )$$

Now consider the nature of $$(\mathbb{Z} / 6 \mathbb{Z} )$$

We have $$0 + \mathbb{Z} / 6 \mathbb{Z}$$ = { ... ... -18, -12, -6, 0 , 6, 12, 18, 24, ... ... }

and $$1 + \mathbb{Z} / 6 \mathbb{Z}$$ = {... ... -17, -11, -5, 1, 7, 13, 19, 25, ... }

and so on

But what is $$3 \mathbb{Z} / 6 \mathbb{Z}$$ ? and indeed, further, what is $$(\mathbb{Z} / 6 \mathbb{Z} ) / (3 \mathbb{Z} / 6 \mathbb{Z} )$$ ?

Can someone please help clarify this matter?

Peter

 

[Deveno]

Let's just consider Z-modules...that is, abelian groups. We can ask:

What does the group (Z/mZ)/(Z/nZ) look like?

Let's consider a more general question: for abelian groups G,H,K with K a subgroup of H, and H a subgroup of G, what does:

(G/K)/(H/K) look like?

the elements of G/K and H/K both look very similar: they are of the form g+K or h+K (additive cosets or translates of K). So a "typical" element of (G/K)/(H/K) is a coset:

(g+K) + (H/K).

Let's examine this in more detail when G = Z, H = 3Z, and K = 6Z.

The elements of Z/6Z are:

{...-6,0,6,12...} = 0+6Z = 6Z = [0]
{...-5,1,7,13...} = 1+6Z = [1]
{...-4,2,8,14...} = 2+6Z = [2]
{...-3,3,9,15...} = 3+6Z = [3]
{...-2,4,10,16...} = 4+6Z = [4]
{...-1,5,11,17...} = 5+6Z = [5]

this is a cyclic group of order 6.

What is H/K = 3Z/6Z? This consists of those cosets whose representatives are all multiples of 3. This is the subgroup of Z/6Z generated by [3], namely:

H/K = {[0],[3]}

in other words the natural projection from G/K to (G/K)/(H/K) sends:

[0]-->H/K
[1]-->[1] + H/K = {[1],[4]}
[2]-->[2] + H/K = {[2],[5]}
[3]-->H/K
[4]-->[1] + H/K (which equal [4] + H/K)
[5]-->[2] + H/K

Clearly, (G/K)/(H/K) has order 6/2 = 3, we have the isomorphism from Z/3Z given by:

[k] = k + 3Z <--> [k] + H/K

In fact, (G/K)/(H/K) is isomorphic to G/H, a fact known as the third (or sometimes second) isomorphism theorem for abelian groups (aka the "freshman theorem"...just "cancel the K's"), via the map:

(g+K)(H/K) <--> g+H

the only "tricky part" here is verifying this map is *well-defined*, that is:

if (g+K)(H/K) = (g'+K)(H/K), then g+H = g'+H. But see:

if (g+K)(H/K) = (g'+K)(H/K), this means that:

(g+K) - (g'+K) = (g - g') + K is in H/K, which means that g - g' is in H.

 

[Math Amateur]

Let's just consider Z-modules...that is, abelian groups. We can ask:

What does the group (Z/mZ)/(Z/nZ) look like?

Let's consider a more general question: for abelian groups G,H,K with K a subgroup of H, and H a subgroup of G, what does:

(G/K)/(H/K) look like?

the elements of G/K and H/K both look very similar: they are of the form g+K or h+K (additive cosets or translates of K). So a "typical" element of (G/K)/(H/K) is a coset:

(g+K) + (H/K).

Let's examine this in more detail when G = Z, H = 3Z, and K = 6Z.

The elements of Z/6Z are:

{...-6,0,6,12...} = 0+6Z = 6Z = [0]
{...-5,1,7,13...} = 1+6Z = [1]
{...-4,2,8,14...} = 2+6Z = [2]
{...-3,3,9,15...} = 3+6Z = [3]
{...-2,4,10,16...} = 4+6Z = [4]
{...-1,5,11,17...} = 5+6Z = [5]

this is a cyclic group of order 6.

What is H/K = 3Z/6Z? This consists of those cosets whose representatives are all multiples of 3. This is the subgroup of Z/6Z generated by [3], namely:

H/K = {[0],[3]}

in other words the natural projection from G/K to (G/K)/(H/K) sends:

[0]-->H/K
[1]-->[1] + H/K = {[1],[4]}
[2]-->[2] + H/K = {[2],[5]}
[3]-->H/K
[4]-->[1] + H/K (which equal [4] + H/K)
[5]-->[2] + H/K

Clearly, (G/K)/(H/K) has order 6/2 = 3, we have the isomorphism from Z/3Z given by:

[k] = k + 3Z <--> [k] + H/K

In fact, (G/K)/(H/K) is isomorphic to G/H, a fact known as the third (or sometimes second) isomorphism theorem for abelian groups (aka the "freshman theorem"...just "cancel the K's"), via the map:

(g+K)(H/K) <--> g+H

the only "tricky part" here is verifying this map is *well-defined*, that is:

if (g+K)(H/K) = (g'+K)(H/K), then g+H = g'+H. But see:

if (g+K)(H/K) = (g'+K)(H/K), this means that:

(g+K) - (g'+K) = (g - g') + K is in H/K, which means that g - g' is in H.

Thank you Deveno, that was a VERY helpful post ... ...