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

[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)