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

[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)