(Meant for my Intro to Homological Algebra Summer Seminar Group at SJC)

 

Last time we met we discussed groups, where we can recall that

A group, G, is a set with a binary operation \mu: G \times G \to G which is associative, has an identity, and has inverse elements.

Example: \mathbb{Z} / 4 \mathbb{Z}

 

We’re going to next consider how to take a group, G and a subgroup, H and form the quotient group G/H.  As an example, let’s use the group (\mathbb{Z}, +) of integers under addition.  Consider the subgroup 4 \mathbb{Z}, which contains elements like 12, 4, -100, etc, but not 3, 2, -99, etc.  We will now define an equivalence relation on \mathbb{Z} by saying that the elements x and y are equivalent, which we will denote by x \sim y, if x - y \in 4\mathbb{Z}.

Let’s go through some examples of elements which are equivalent:

  • 4 - 0 = 4 \in 4 \mathbb{Z} and so 4 \sim 0. [Edited 6/13]
  • 4 - 8 = -4 \in 4 \mathbb{Z} and so 4 \sim -8.
  • 4 - 5 = -1 \notin 4 \mathbb{Z} and so 4 \not\sim 5.

If we went through all possible combinations, we would see that there are exactly 4 equivalence classes of elements that we are partitioning \mathbb{Z} into:

  • 4 \mathbb{Z}+ 0 = \{\ldots, -12, -8, -4, 0, 4, 8, \ldots \}
  • 4 \mathbb{Z}+ 1 = \{\ldots, -11, -7, -3, 1, 5, 9, \ldots \}
  • 4 \mathbb{Z}+ 2 = \{\ldots, -10, -6, -2, 2, 6, 10, \ldots \}
  • 4 \mathbb{Z}+ 3 = \{\ldots, -9, -5, -1, 3, 7, 11, \ldots \}

Just to absolutely clear, for each of these sets, if I take two elements in the same set, and I look at their difference, it will be a multiple of 4; if I take two elements each from different sets, their difference will not be a multiple of 4.

Here’s where things get trippy: we will now take the infinitely-large set, \mathbb{Z}, and just treat it as four elements: [0] = 4 \mathbb{Z}+ 0 [1] = 4 \mathbb{Z}+ 1 [2] = 4 \mathbb{Z}+ 2 , and [3] = 4 \mathbb{Z}+ 3 .  Note: a shortcut for knowing which set is which is by looking at the remainders of the elements, so 5 is in [1] because when I divide it by 4, the remainder is 1.  And -3 is in the [1] as well because when I consider: (-3) – (5)= – 8, I get an element in 4 \mathbb{Z}!

 

So now let’s add elements in our set, \mathbb{Z} / 4 \mathbb{Z} = \{ [0], [1], [2], [3] \}.  Remember that each [x] is a subset of integers.  Suppose I took the number 2 \in [2] and added it to the element 3 \in [3].  Well 2+3 = 5, but there is no element “[5]“, is there?  Well if there was, what would it be?  The set [5] would be the set of integers which are equivalent to the number 5, under the rule “~” we used above.  Well note that 5 \in [1], and so it turns out that [5]=[1] (by the properties of an equivalence relation, which we will have to go through carefully and formally).  The point is… [2] + [3] = [5] =[1]!!

 

Example: \mathbb{Z} / 5 \mathbb{Z}

Let us do the same thing as we did above to partition the integers, except now x \sim y if x -y \in 5 \mathbb{Z}.  Note that we now have 5 elements in our set, which the rest of the world refers to as \mathbb{Z}_5 := \{ [1], [2], [3], [4], [5]\}.  Note that I chose to use [5] in place of [0]; there is a reason which you might come to realize but it isn’t super important.

While the integers only form a group under addition, you should explore this new set and see if it forms a group under addition or multiplication or both!?  Either way the answer should be surprising and interesting.

 

Example: 0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0

Finally, I would like to introduce a more geometric example.  The previous examples ended up with a finite number of elements in our quotient, and everything was down to arithmetic.  Consider again our lovely set of integers, \mathbb{Z}, which are a group under addition.  But now consider the subgroup 2 \pi \mathbb{Z} \subset \mathbb{R} of integer-spaced multiples of 2 \pi, each of which is a real number (whatever that means, amiright?).  Anyway, I’m now going to consider the group \mathbb{R} / 2 \pi \mathbb{Z} , but what the heck is this group?

Well I chose 2 \pi because that is the circumference of the unit circle.  So notice I have a function f: \mathbb{R} \to \mathbb{R}^2 by sending t \mapsto (\cos (t), \sin (t)) which takes each real number and maps it on the unit circle.  Which real numbers go to zero?  The multiples of 2 \pi!  Which real numbers get sent to \frac{\pi}{3}?  The real numbers 2 \pi \cdot n + \frac{\pi}{3} where n is any integer!

So we’ve taken the real line, and thought of any two numbers which are 2\pi-apart as being glued together, and if you try to draw this picture you should see a circle!

 

cycloid

Well this picture isn’t really what I want you to see but I’m tired and it should be good enough to get your creative brain going!

 

Food For Thought: Preparing for our next meeting

After having slowly read through this blog post, which could take hours, I encourage you to try and go back through our notes to see if any of the examples, propositions, exercises, etc, make any more sense than they did.  Specifically, I want us to finish up this discussion on quotients in our next meeting and then continue on to some other ideas, which I will now appropriately tease…

Note that we started with the additive group of integers, and when modding out by 5 \mathbb{Z}, we actually get some nice multiplicative structure.  What if a group had both an additive and a multiplicative structure, do quotients make sense for playing nicely with both of these structures simultaneously?  The answer to this leads us to a study of rings (groups which also have some multiplicative structure; not necessarily inverses), and modules (groups which can be multiplied by external elements; like vectors can be multiplied by numbers).

Finally, in the last example I gave, the title for that example had some arrows connecting the integers, to the real numbers, and then to the circle, S^1.  Those arrows are actually functions which allow the group structures of each group to communicate with each other.  This type of function is called a homomorphism and will be central to our study of Homological Algebra.

2 thoughts on “Quotients of Groups: Where 2+3 = 1.

  1. Thanks for the content, but I think you have a typo! Where you specify examples of which two elements are equivalent, your third example shows that 4 is clearly not equivalent to 5, but your first example shows that 4 is equivalent to 5…! I noticed it as soon as I realized that 5 is clearly not a multiple of 4, so maybe you meant to say that 4 ~ 4.

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