Arithmeticae. Here he laid out the modern idea of modular arithmetic. This technique

takes the group of integers, partitions them into a finite number of sets, and then treats

each set as a new type of number. Modular arithmetic has proven to be so useful, it’s

natural to try to adapt this technique to other areas. In group theory, this can been

done using normal subgroups and quotient groups.

To motivate the idea of a normal subgroup and a quotient group, let’s first look at

a concrete example. Let’s begin by looking at the integers mod

5. Here, we will divide the integers into 5 sets,

depending on the remainder you get when you divide the number by 5.

The first set will be all the integers with a remainder of 0 when you divide by 5. In

other words, all the multiples of 5 - positive, negative, and zero.

The next set is the integers with a remainder of 1 when you divide by 5.

We continue this process: those with a remainder of two… three… and four… We can now

stop, because these are the only possible remainders you can get.

Here’s the critical observation: if you pick any number with a remainder of 1, and

add it to any number with a remainder of 2, you will always get an integer with a remainder

of 3… Similarly, if you pick any number from the

2 set, and add it to any number from the 4 set, you will always get a number in the 1

set... This always happens. If you pick any number

from one set, and add it to any number from a second set, the sum will always be in the

same, third set.

[4] These five sets are called congruence classes, and if we treat the sets as if they

were numbers, then we have a group with five elements: the integers mod 5. It’s very

good practice to check that these 5 “meta numbers” do indeed form a group under addition.

For instance, the set of the multiples of five act as the identity element, and each

set has an inverse. By the way, whenever two integers “A” and “B” are in the same

congruence class, we write it like this. We say this aloud as “A is congruent to B mod

N”, and all this means is that “A” and “B” have the same remainder when you divide

by N.

Let’s take another look at the integers mod 5 using the language of group theory.

To start, we have the group of integers “Z” under addition.

This group has an infinite number of subgroups, but we’ll look at the subgroup of the multiples

of 5, which we’ll write as “5-Z.” Graphically, let’s represent the group of

integers by a rectangle, and the subgroup “5-Z” as a smaller rectangle inside of

it. Every integer in this subgroup has a remainder

of 0 when you divide by 5. The number 1 is not in the subgroup, and if

you add 1 to every number in the subgroup, you get a new set -- the set of integers with

a remainder of 1 when you divide by 5. We’ll denote this set by 1 + 5Z.

We call this set a “coset” and it’s NOT a subgroup. It’s not closed under addition,

doesn’t have inverses, and does not contain the identity element. It’s not even close

to being a group. This coset does not overlap with the subgroup

5-Z, since every number in the coset has a remainder of 1 when you divide by 5, and every

number in the subgroup has a remainder of 0 when you divide by 5.

We can continue this process by picking a number not already inside a rectangle, like

2, and then making a new coset: 2 + 5Z. Continuing this process we get the cosets

3 + 5Z… and then 4 + 5Z... At this point, the original group is now completely

covered by one subgroup 5Z, and 4 cosets. By the way, you can also think of the subgroup

as a coset: 0 + 5Z.

We used the subgroup 5-Z to partition the group Z into cosets. Because the cosets form

a group, we call 5-Z a NORMAL SUBGROUP. And the group of cosets is called a QUOTIENT GROUP,

and it’s written like this. This name is very descriptive, since we are using a subgroup

to divide the group into cosets. And when you divide one thing by another, you get a

quotient. But the critical insight in this process was the observation that you can treat

these cosets as elements in a new group: a coset group, if you will.

For example, if you add the coset 1 + 5Z to the coset 3 + 5Z you get the coset 4 + 5Z.

The way you add these two cosets is to add all the numbers in the first coset with all

the numbers in the second coset. The resulting set turns out to be exactly the coset

4 + 5Z.

I’d like to point out one more thing. In the example we’ve discussed, we started

out with a group Z, then took a subgroup 5Z -- which is called a normal subgroup -- and

used these two to create a quotient group out of the cosets. The quotient group is NOT

a subgroup of Z. It’s an entirely different group.

Let’s now begin the process of generalizing this technique to an arbitrary group. Suppose

we have a group G, and a subgroup N. Here, we will use multiplicative notation for the

group G. Like before, we can use N to generate a collection of non-overlapping cosets. Remember,

N is a subgroup, while the other cosets are simply sets. Here is the big question: do

the cosets always form a group? The answer is NO. If the cosets do not form a group,

we do NOT call N a normal subgroup, and we CANNOT make a quotient group. Let’s now

see what properties N must have in order for the cosets to be a group.

Let’s assume N divides G into T different cosets.

Since G may not be abelian, we need to be careful: left cosets and rights cosets may

be different, so we’ll go ahead and work with left cosets.

Now, every left coset is of the form g-N for some element ‘g’ in the group.

Let’s pick two different cosets: X-N and Y-N.

Since N is a subgroup, it contains the identity element ‘E’.

This means X times E (which equals X) is in the first coset, and Y times E (which equals

Y) is in the second coset. So if the cosets behave like a group, X times

Y must be in the product of the two cosets. In other words, X-N times Y-N should equal

XY-N. If this is true, then the product of any element

in the first coset with any element in the second coset should be in the coset XY-N.

Let’s see when this happens. Pick an arbitrary element in the first coset,

call it X-times-N1. And then pick an element from the second coset,

Y-times-N2. If we multiply these together we get X-N1

✕ Y-N2. If this is in the coset XY-N, then their product

must be XY-times-N3 for some N3. We’re now going to tinker with this equation

to get a simpler expression. We can simplify this by multiplying on the

left hand side by the inverse of X and cancelling. Next, multiply on the left by Y-inverse and

cancel on the right. This gives us Y-inverse ✕ N1 ✕ Y ✕ N2

EQUALS N3. Finally, multiply this on the right by N2-inverse.

Since N is a subgroup, the product on the right is also in N.

So we end up with Y-inverse ✕ N1 ✕ Y is an element of N.

In fact, if you look at the set Y-inverse ✕ N ✕ Y you get N.

One way to do this is to show each side is a subset of the other. We just showed the

left hand side is a subset of N, and it’s a very good exercise to show that the right

hand side is a subset of the left.

So to summarize, if the product of the two cosets X-N and Y-N is well defined, then it

must be true that Y-inverse ✕ N ✕ Y equals N. We call the left hand side a conjugate.

Now that we have a group operation, we can check that the cosets form a group.

The identity element is just the subgroup N, which you can also think of as the coset

E-N. We see this is the identity because E-N times

G-N equals (E✕G) N which equals G-N. And for each coset G-N the inverse is G-inverse-N,

because if you multiply these two cosets together, you get (G✕G-inverse) N which equals E-N,

or simply N, which is the identity element.

We just saw that if N is a subgroup of G and the cosets behave like a group, then it must

be true that the set Y-inverse TIMES N times Y must equal N for any Y. But with a clever

trick, we can show the converse is true as well. That is, if the conjugate of N always

equals N, then the cosets form a group. Let’s see why.

Suppose Y-inverse ✕ N ✕ Y equals N for any element Y.

We’ll use this fact to show that the cosets form a group.

To begin, pick two cosets X-N and Y-N. Let’s multiply two arbitrary elements from

these cosets: X-N1 and Y-N2. Here’s the clever trick: Y ✕ Y-inverse

equals the identity element E. Multiplying by E has no effect, so let’s

insert Y ✕ Y-inverse right after X. In the middle of this expression is Y-inverse

✕ N1 ✕ Y. But from our assumption, we know this must

be an element of the subgroup N, call it N3. So this expression equals X times Y times

N3 times N2. But N3 times N2 is also in N, call it N4.

So the expression further simplifies to X ✕ Y ✕ N4 which is an element of the coset

X-Y-N. This means the product of the cosets X-N and

Y-N is the coset X-Y-N. The cosets do form a group.

What we’ve just shown is that if N is a subgroup of G, then the cosets behave like

a group precisely when Y-inverse TIMES N TIMES Y equals N for any element Y in the group.

When this is true, we call N a “normal subgroup” of G, and we write it like this. The group

of cosets is called a “factor group,” and it’s written like this… In the factor

group, the subgroup N is the identity element, and the inverse of X-N is X-inverse-N.

Every group G has two subgroups: the identity element and the entire group G. It turns out

these are technically normal subgroups, but they aren’t very interesting. If a group

has no other normal subgroups than these two, then we call G a SIMPLE GROUP. A simple group

does not have any factor groups, and they are the building blocks of other groups, much

like prime numbers are the building blocks of the integers.

Normal subgroups and quotient groups are among the most useful devices in abstract algebra.

In separate videos, we’ll show how normal subgroups determine what kinds of homomorphisms

are possible from a group G to other groups. And for finite groups, you can find a chain

of normal subgroups called a “composition series” which acts as a kind of “prime

factorization” of the group. Normal subgroups can even be used to study fields; you’ll

learn about this technique in Galois Theory.

Here’s a good exercise. Try to find a normal subgroup of the symmetric group S3. This group

has only 6 elements, so you should be able to check everything by hand. Be sure to help

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