Is Ker(f)={e} Enough to Prove Injectivity for a Group Homomorphism?

[matt grime]

no, post 32 was about isomorphisms. think about it. if g is an element of any group and g^n=e what does that tell you about the order of g?

from what you;ve said you appera to think that there is a non-zero homomorphism from Z_7 to Z, and there isn't.

 

[quasar987]

The order of g divides n.

How is f([m]) = a^m not an homomorphism from Z_7 to Z?

 

[matt grime]

What is a? why does the order m, which divides 7, divide the order of a^m? how can it? there are no elements of finite order in Z apart from 0 are there? so the only homo from Z_7 to Z is the trivial one.

 

[matt grime]

one thing that would really help you is to learn the isomorphism theorems (and don'r sey they are in another chapter. one can prove all these things without tem but to understand why they are true it is ehlpful to know them).

if f is a homo from G to H it means that there is a quotient group of G isomorphically embedded in or has a copy inside or is a subgroup of H. Z_7 has no non-trivial quotients os it's clear that theer can be no isomorphic embedding of Z_7 in Z, or less fancilly, Z_7 is not a subgroup of Z, though you have stated it is.

 

[quasar987]

a is some element in Z.

Defined in this way, f([m]+[l]) = f([m]) f([l]), so f is an homomorphism. That's how I see it. How do you?

 

[Muzza]

But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...

 

[matt grime]

oh come on! I've already explained it isn't a homomorphism unless a=0 at which point it is trivial. in the sepciifc example you gave you were asked to send 1 in Z_7 to g in G where g=/=e so that rules out the trivial case.


Let is use addition since Z is an additive group, as is Z_7

you are sending [m] in Z_7 to am in Z

this isn't a homomorphism, and you should prove it. HINT 3+4=0 in Z_7 and doesn't in Z

 

[quasar987]

matt, let's start with the first problem, which is

If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from Z_n to G which sends [1] to a? What are the homomorphisms from Z_2 to Z_6?

But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...

Oh, so the problem is that f is not even a function! Interesting... Well it would be if a is e!

Ok, so now to the first question, I can answer that f, defined as

$f([m])=a^m$

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a $\neq$ e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.

To the other, I would say that the homomorphisms from Z_2 to Z_6 are

$$\{f_i([m]) = 3i[m]\}_{i \in \mathbb{Z}}$$

 

[matt grime]

you appear to have an infinitely family of homomorphisms, are you claiming that they are different?

Suppose the map from Z_2 to Z is an isomorphism, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3. If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.

try it if we think of maps Z_3 to Z_6

 

[quasar987]

No, I don't claim that; the f_i's for odd i's all do the same thing, namely send [m] to [3] and the f_i's for even i's all do the same thing, namely send [m] to [0].

Let me now ponder on your isomorphism argument there.

 

[quasar987]

I can't crack the code :tongue:

Suppose the map from Z_2 to Z is an isomorphism

The map? What map? Also, is it not impossible to have an isomorphism btw a group of order 2 and one of infinite order?

, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3.

Please use the [] notation for elements of Z_n in order to avoid confusion, thx.

If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.

Two homos! Could you extrapolate the logical steps leading to that conclusion?

 

[matt grime]

apologies, i meant to say "suppose f is an injective homomorphism, ie its kernel is {e}, from Z_2 to Z_6" so there were at least 2 errors there.

no, i won;t use brackets for Z_n since it is unnecessary.

 

[quasar987]

Ok, so now to the first question, I can answer that f, defined as

$f([m])=a^m$

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a $\neq$ e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.

How do we show that all homomorphisms that send [1] to a are of the form above?

edit: Or rather, how do we show that f above is the only homo that sends [1] to a.

 

[AKG]

$\mathbb{Z}_n$ is cyclic, so to find what a homomorphism does to any element m in Z_n, you only need to know what it does to 1, since 1 generates the group:

f(m) = f(1^m)
= (f(1))^m since f is a homomorphism
= a^m since we stipulated that f sends 1 to a

Note that normally, a^b means that we multiply a, b times. However, when dealing with groups, it is common to refer to the group operation as multiplication (regardless of what it is - addition, addition mod n, multiplication, composition, etc.) and it is common to use "multiplicative notation," regardless of what the operation is. So 1^m means that we do the group operation to 1, m times. The group operation happens to be addition modulo n, so we might prefer to write m(1) instead of 1^m, but as long as you keep in mind that there is only one operation that concerns us, then the notation 1^m meaning to add 1 to itself m times shouldn't confuse you.

Anyways, now you should be able to see that f must send m to a^m. There is only really one restriction that needs to be placed in order for f to be a function. Once this restriction is in place, not only will f be a function, but by the way we defined it, it will automatically be a homomorphism. Can you figure out this restriction? n need not divide the order of G. In fact, G could be an infinite group, in which case it may not even make sense to say whether n divides |G| or not.

 

[quasar987]

It must be that the order of a divides n: o(a)|n

 

[AKG]

Correct. Equivalently, aⁿ = e.

 

[quasar987]

Yes. Well that was easy afterall ..but then again, everything seems easy afterwards.

 

[matt grime]

the eternal difficulry of mathematical teaching. where are you learning this from? since any book or course should surely go:

Define a group
Prove lagrange's theorem.
Define a homomorphism
Define an isomorphism

prove that if x has order n then f(x) has order dividing n (automatically this shows there is only ever 1 homomorphism from a finite group to Z, or any group where there are no elements of finite order except for the identity. such groups are called torsion free and are very significant)

prove that isomorphisms do respect "all algebraic properties" or at least prove lots of cases. for instance can you prove that if f is an iso from G to H that

x and f(x) have the same order
y is conjugate to x iff f(y) is conjugate to f(x)
y commutes with x if and only if f(x) commutes with f(y)

prove that if G is generated by a set of elements S any homomorphism is determined by where it sends S

 

[quasar987]

As I said in the other thread (Group question), I'm using this book: http://www.math.miami.edu/~ec/book/

It does not cover Lagrange theorem as far as I know.

edit: Oh yes, it does; it is camouflaged as the fifth statement of a bigger theorem at page 25 (or 33 according to adobe reader).

 

[AKG]

Define a group
Prove lagrange's theorem.
Define a homomorphism
Define an isomorphism

My book goes nothing like that, but I think it's a pretty good book.

 

[mathwonk]

that looks like a nice book. and ed connell is certainly very strong mathematician. that book is written concisely so you should spend a lot of time on each page.

 

[matt grime]

surely lagrange is the first thing one learns in group theory? then shortly afterwards the orbit stablilizer theorem and the class equation. what else is there in basic group theory?

 

[mathwonk]

in fact orbit stabilizer and class equation are special cases of the lagrange theorem!

i.e. the orbit is naturally the set of cosets of the nstabilizer, of any given element, so the order of the orbit times the order of th stabilizer equals the oder of th group.

and the class equation is, aha, there is a little more here since we are summing the orders of the orbits over every conjugation class, but it is still just the orbit stabilizer result repeated.

of course you need to know what a homomorphism, i.e. an "action", is to link these to the lagrange theorem. i.e. you need matt's second basic concept

so lagrange is absolutely fundamental as matt said.

then the rest of elementary group theory is an attempt to prove the converse of lagrange. i.e. sylow tells you when, given a number that divides the order of the group, you can find a subgroup of that order.

it is always true at least when the number is a prime power.

and the proof, uses the counting principle encoded in the stabilizer orbit theorem!

 

[AKG]

I would think that before learning a theorem like Lagrange, you should have some familiarity with a variety of groups, right? Permutation groups, number groups, matrix groups, symmetry groups, etc. As for more general theorems, there's Cayley's, Cauchy's, the Sylow theorems, isomorphism theorems. This is the way my book does it:

Symmetries of the Tetrahedron
Axioms
Numbers
Dihedral Groups
Subgroups and Generators
Permutations
Isomorphisms
Plato's Solids and Cayley's Theorem
Matrix Groups
Products
Lagrange's Theorem
Partitions
Cauchy's Theorem
Conjugacy
Quotient Groups
Homomorphisms
Actions, Orbits, and Stabilizers
Counting Orbits
Finite Rotation Groups
The Sylow Theorems
Finitely Generated Abelian Groups
Row and Column Operations
Automorphisms
The Euclidean Group
Lattices and Point Groups
Wallpaper Patterns
Free Groups and Presentations
Trees and the Nielsen-Schreier Theorem

"Groups and Symmetry", M. A. Armstrong

 

[mathwonk]

you can make anything look as complicated as you wish. but matt is giving you the essence of the theory. of course examples flesh out the theory.

at least half your list is examples.

 

[mathwonk]

and by the way, "lagrange's" theorem is originally due to gauss in the case of integers mod n, in his disquisitiones.

 

[AKG]

of course examples flesh out the theory.

at least half your list is examples.

I think that's the point. Learning the theory is much easier when you are familiar with a variety of more concrete examples. It also allows for more interesting and varied problems. If you don't "flesh" out the theory, there's little for a student to "grab" on to, at least that's what I've found.

 

[mathwonk]

i agree those are essential, but it also helps to subdivide the chapter headings according to what is contained in them.

you should realize some of it is examples of what has come before, in order to grasp the structure of the subject.

 

[matt grime]

familiar with concrete examples is good. indeed the notes i have written on group theory are predicatd on the idea that concrete groups are easier to understand. but knowing what an example of a group is doesn't help you to manipulate them.

http://www.maths.bris.ac.uk/~maxmg/docs/groups.pdf

uses the examples to discuss special cases of orbit stabilizer so that they can see what the proof id going to do. it doesn't prove lagrange as it was written for an audience that was already familiar with lagrange.

for me lagrange is the essential tool to understand the examples - why isn't there a non-trivial homo from Z_2 to Z, why if I'm finding the order of an element in G a group of order 6 do i need onyl to check the 2nd and 3rd powers? why is the 6th power always the identity? certainly give some examples but to exploer the examples we ought to have lagrange.

sadly the course i was tutoring on groups came before the students were (officially) introduced to matrices, though some had done them at school. but that#s a whole new thread on falling standards again.

 

[quasar987]

Excuse me to interupt, but it seems there's still a missing piece to solving the 2nd problem (what are the homos from Z_2 to Z_6). I've found that there is a homo that sends [1] to [a] iff o([a])|2 and in such case, the homo is f([m])=m[a].

So I can check the order of every element [a] of Z_6 and see if o([a])|2 for that element. If yes, then f([m])=m[a] is an homo, if not, it is not a function. But that allows me only to find all the homos from Z_2 to Z_6 that send [1] to [a], i.e. the homos of the form f([m])=m[a]. I still don't know if there are other possible homos that are not of the form f([m])=m[a].

 

[matt grime]

we already told you that since Z_2 is generated by 1 it only matters what 1 is sent to, indeed i said you should prove that the image of a homomorphism is determined by where it sends generators. moreover there is another reason: since Z_2 contains exactly two elements, 0 and 1, and 0 must be sent to the identity you only need to work out where 1 goes.

i thought we established that there are exactly two homos from Z_2 to Z_6

0 --> 0
1--> 0

and
0-->0
1-->3

 

[quasar987]

Oh, right! Where [1] is sent too automatically determines where everything else is sent. Double sweet. Thanks you guys.

 

[matt grime]

That is automatically true for Z_m for any m (and only for groups isomorphic to Z_m since they completely classify the (finite) groups with a sinlge generator) and tis is doubly true for Z_2 since not only is 1 a generator but it is the only non-identity element. i think (you really need to stop using [1] because almost no one else does so it is good to get into these bad habits, but then that is just what i think)