How to Define Homomorphisms on Quotient Groups

so we are going to talk about defining

homomorphisms on quotients this video is

going to use the language of groups but

the same idea applies to other algebraic

structures such as rings

so let's say we have a group G and a

normal subgroup K of G we can form the

quotient Group G mod K and the elements

of that quotient are cosets which look

like G times K where little G is an

element of the original group

now let's say we want to define a

homomorphism f that goes from the

quotient Group G mod K to some other

group h

one way we can Define such a

homomorphism is to say well F the

elements of the quotient group The

Domain of f

those are these cosets G times k

and when we take a coset this element

here this little G is called a

representative of the coset G times K

it's a specific element of the coset

that we choose to represent the entire

set

so one way to define F of G times K is

to just look at the coset representative

and Define f as a function of the

representative

so to do this we're essentially defining

another homomorphism fee that goes from

the original group G to that same

codomain H and then we're using this fee

to Define f in terms of the coset

representatives

now if we Define f in this way if Phi is

a group homomorphism then F defined like

this is automatically also a group

homomorphism because if we look at F of

g 1 K times G2K where G1 and G2 are

elements of that group

in here this g1k times G2K is the same

as G1 G2K that's how multiplication in

the quotient group is defined

and then

by definition F of G1 G2K that's going

to be Phi of G1 G2

and then on the other hand F of g1k

times F of G2K

well we can compute each of these

individually using this definition so F

of g 1 K is Phi of G1 and similarly F of

G2K is Phi of G2 now because Phi is a

group homomorphism Phi of G1 G2 equals

Phi of G1 times Phi of G2 which means

that these two first expressions are

equal and therefore f is a group

homomorphism

so it seems like we're done if we want

to define a homomorphism on the quotient

group gmod K all we have to do is find a

homomorphism on the original group G and

then use that to define a function on

the cosets by just picking a coset

representative and plugging that into

Phi

there's one problem here though which is

we don't know whether the function f is

well defined

to see an example of where this problem

comes from let's take a look at the

group of integers under addition

we can define a group homomorphism from

the integers to Itself by just taking

Phi of n equals n this is the identity

map on the integers of course it's a

group homomorphism

now if we take a look at the subgroup 5z

this is a normal subgroup of the entire

group of integers Z under addition so

this subgroup here is just the set of

all integers that are multiples of five

and we can construct the quotient group

Z mod 5z so this is the integers mod 5.

and using this construction we could

Define a function f that goes from Z mod

5z to the same codomain Z

and the way we want to Define that is by

saying F of n plus 5z this is our coset

and the definition here is that we just

take the coset representative and apply

Phi to that so Phi of n

but Phi of n is just equal to n

so take a second and see if you can

figure out the problem with this

definition of a function f of n plus 5z

equals n

the problem here is that

n plus 5z is a coset that contains many

different elements so as an example

let's take a look at the coset 2 plus 5z

so this coset contains 2 plus every

integer multiple of five so the elements

are going to look like 2 7 12 17 and

then also in the reverse Direction we'd

have minus three minus 8 minus 13 and so

on

but the key Point here is that 2 plus 5z

is the same coset as 7 plus 5z because

if we take 2 and add all the multiples

of 5 the set that we get is the same set

that we would get if we start with 7 and

then add all the multiples of five

but if these are the exact same set then

we need to have F of 2 plus 5z being

equal to F of seven plus five Z

and that's not what we get with this

definition because F of 2 plus 5z is

just 2 and F of 7 plus 5z is seven but

that doesn't make any sense because

these 2 plus 5z and 7 plus 5z they're

the same set so we can't have a function

that Maps the same set to two different

outputs

so let's take this problem back to the

more General language of an arbitrary

group

the problem here is that if we have some

coset GK of a normal subgroup k then we

can also have a different coset

representative let's say the first

representative is G1 and the second

representative is G2 so we're almost

always going to have cases in a quotient

group where there are two elements G1

and G2 of that original group where g1k

equals G2K and if that's the case then

we need F of g1k to equal F of G2K

if we want to define a homomorphism on

the quotient group in terms of

homomorphism on the original group G

then we're having F of G1 K being equal

to Phi of G1 and then F of G2K is Phi of

G2

so in order for the function f to be

well defined we need to have that if g1k

equals G2K then Phi of G1 equals Phi of

G2 so let's think a little bit more

about what this equation here means if

g1k equals G2K one thing we can do here

is multiply both sides of the equation

by G2 inverse on the left

so on the left side we'll get G2 inverse

G1 K and then on the right side G2

inverse G2K is just k

now both sides of these equation these

are sets of elements in the original

group G so of course this K on the right

side is just that normal subgroup but on

the left side G2 inverse g1k that's the

set of all elements of the form G2

inverse times G1 times some element of

the normal subgroup k

now in particular K contains the

identity Element e of the original group

so G2 inverse G1 times the identity

element

this is in G2 inverse G1 K this is part

of that coset

and we know that G2 inverse g1k equals

that normal subgroup k

which means that G2 inverse G1 times the

identity is an element

of K of course the identity element

doesn't do anything when we multiply so

this is telling us that G2 inverse G1 is

an element of K so this first equation

g1k equals G2K

that means that G2 inverse G1 is in that

normal subgroup k

now let's take a look at the second

equation over here Phi of G1 equals Phi

of G2

now in this case Phi of G1 and Phi of G2

these are both elements of the second

group h

so what I'm going to do is multiply by

the inverse of this element Phi of G2 on

both sides

so on the left side we're going to get

Phi of G2 inverse times Phi of G1 and

then on the right side Phi of G2 inverse

times Phi of G2 that's just the identity

element

but fee is a group homomorphism so first

of all Phi of G2 and then taking the

inverse that's the same as Phi of G2

inverse

and then second of all Phi of G2 inverse

times Phi of G1 is the same as Phi of G2

inverse G1 so those are just basic

properties of the group homomorphism

so Phi of G2 inverse G1 equals the

identity that's what this second

equation is telling us

so the new form we have of the

implication that we're looking for here

is that if G2 inverse G1 is in k then

Phi of G2 inverse G1 equals the identity

but notice in both cases here that this

element here G2 inverse G1 that's the

same as what we're plugging into Phi

so what we're really saying is that if

we have some element of the group G

that's in the normal subgroup k then Phi

has to be the identity on that element

of the group that's the thing that we

want in order for the homomorphism to be

well defined

so in other words fee applied to any

element of the normal subgroup K has to

be the identity

so this is the condition that we need to

check whether a homomorphism on the

original group G gives us a well-defined

homomorphism on the quotient Group G mod

k

we start with any homomorphism fee from

G to H and all we need to check is that

Phi gives us the identity on every

element of that subgroup k

as an example of this condition we can

take a look at the homomorphism fee from

the integers to itself given by Phi of n

equals n

in this case we're considering the

normal subgroup 5z that's basically our

K in this case and one element of 5z

that's all the multiples of 5 is just

the integer five but Phi of five equals

five and that's not the identity element

of the integers it's not zero so this

homomorphism fee does not satisfy this

condition which is why it doesn't give

us a well-defined homomorphism on the

quotient group Z mod 5z

next I'll give an example of this

condition working if you're familiar

with the symmetric group and the sign of

a permutation

so there's a group homomorphism Sigma

that goes from the symmetric group SN to

the set plus or minus one this set is a

group under multiplication and the

homomorphism is defined by Sigma of some

permutation g equals the sine of G

so the sign of a permutation tells us

the number of two cycles that are needed

to create the permutation G so if the

number of two cycles that we need to

create the permutation is even then the

sine of G is plus one and we say that g

is an even permutation whereas if the

number of two cycles that we need is odd

then we say that g is an odd permutation

and its sine is negative one so for

example the permutation one two three

this is a three cycle but we can write

it as 1 2 times 2 3. so both one two and

two three are two cycles so this is

writing the permutation as an even

number of two cycles which means that

this is an even permutation

now

the set of even permutations in the

symmetric group actually forms a normal

subgroup it's called the alternating

group a n

so it's a subgroup of index two

and the point is that the sine of every

element in the alternating group is plus

one because by definition this is the

set of all even permutations so it's the

set of all things with sine one and plus

one is the identity element of this

group here

so every element of this normal subgroup

gets sent to the identity by this map

Sigma which means that we can use Sigma

to define a map F on the quotient SN mod

a n to the group plus or minus one if we

Define f on some permutation G times a n

this is a coset of the alternating group

if we Define this as just Sigma of that

permutation G then this gives us a

well-defined homomorphism from SN mod a

n to the group plus or minus one

now the reason that this idea of

defining homomorphisms on a quotient is

so important is that there are many

important objects in math that are best

defined as quotient groups quotient

spaces quotient modules or something

like that

for example if you're familiar with the

tensor product of vector spaces that is

something that we often construct as a

quotient space of a very big Vector

space and then we take the quotient by a

Subspace generated by a bunch of

different elements

so if we Define the tensor product as

this quotient let's say a mod b

it can be very difficult to check

whether a map from a mod b to some other

Vector space is well defined because we

don't know whether it's independent of

the coset representative that we choose

here

so a lot of the time instead of defining

a map directly on the tensor product

what we do is start by defining a map on

the original group a and then we verify

that that map is zero on every element

of this Subspace B that we use to take

the quotient because if the map on a is

zero on every element of B then this

automatically tells us that we get a

well-defined map from a mod b to

whatever other Vector space we're

interested in

so this is a key way that we can Define

homomorphisms on quotients

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