You are honestly claiming that you don't see an isomorphism between and (here is the unit of )?
There is an example on pg-155-156 of Pinter's abstract algebra book which is:
The Problem:
Let and be any groups and consider their product .
Remember that consists of all the ordered pairs as ranges over and ranges over . You multiply ordered pairs by multiplying corresponding
components; that is, the operation on is given by
Now, let be the function from onto given by
It is easy to check that is a homomorphism. Furthermore, is in the kernel of if and only if . This means that the kernel of consists of all the ordered pairs whose
second component is . Call this kernel ; then
We symbolize all this by writing
By the fundamental homomorphism theorem, we deduce that .
[It is easy to see that is an isomorphic copy of ; thus identifying with , we have shown that, roughly speaking, .]
My question is: why and how is that isomorphic to ?
Can anyone kindly tell me: why's that here isomorphic to ?
Sorry that my analyzing capability is little slow. Here are my problems.
According to the definition of isomorphism(Pinter's AA book on pg-89):
Let and be groups. A bijective function with the property that for any two elements and in ,
is called an isomorphism from to .
If there exists an isomorphism from to , we say that is isomorphic to .
I can't find a way to show that is both one-to-one and onto. I know that because of homomorphism.
In other words how do I prove that:
1) If then and
2) There is a function for every ?
This depends on what f is. How do you define it?
Are you given that f is a homomorphism, or are you trying to prove it? In the second case, you obviously don't know that f is a homomorphism yet.
I feel almost silly offering to consider the map as a potential isomorphism.
I think I'm causing confusion. Suppose there is an example like below:
-----------------------------------------------------------------------------------------------------------------------------------
Let and be any groups and consider their product .
Remember that consists of all the ordered pairs as ranges over and ranges over . You multiply ordered pairs by multiplying corresponding
components; that is, the operation on is given by
Now, let be the function from onto given by
It is easy to check that is a homomorphism. Furthermore, is in the kernel of if and only if . This means that the kernel of consists of all the ordered pairs whose
second component is . Call this kernel ; then
...
-----------------------------------------------------------------------------------------------------------------------------
Now how do I mathematically prove that is isomorphic to ? Is it possible that is isomorphic to ?
But the author(Pinter) said that on pg-156:
"...... is a isomorphic copy of ....." . Then where am I wrong?
consider the mapping:
h:G*--->G given by: h(g,e_{H}) = g.
although it is very simple, we can easily prove h is an isomorphsm of G* with G.
1) h is a homomorphism:
h((g,e_{H})*(g',e_{H})) = h((gg',e_{H})) = gg' = h((g,e_{H}))h(g',e_{H}))
2) h is injective:
suppose that h((g,e_{H})) = h((g',e_{H})). this means that g = g', so (g,e_{H}) = (g',e_{H}) (both coordinates are equal).
3) h is surjective:
suppose g is ANY element of G. then (g,e_{H}) is an element of G* that h maps to g.
****************
it may make more sense to consider the inverse isomorphism h^{-1}:G-->G* given by: h^{-1}(g) = (g,e_{H}). if we extend the co-domain to GxH this is an example of an embedding of G in GxH
(an embedding is a monomorphism from one group into another). even though G* isn't strictly EQUAL to G, as groups they are isomorphic (the second coordinate in G* is always e_{H}, which pretty much sits around and does nothing all day).
we use embeddings all the time, and don't even think about it, here is another example:
the integers and rational numbers are two "different kinds of things". an integer is a series of 1's or -1's:
5 = 1+1+1+1+1
-4 = (-1)+(-1)+(-1)+(-1)
a rational numbers is an equivalence class of pairs of integers:
a/b = {(c,d) in ZxZ*: ad = bc}.
nevertheless, it is common to embed the integers in the rationals by the map n--> [(n,1)] (we think of the integer 5 as 5/1 = 10/2 = 15/3 =....etc.)
*********************
with the direct product group GxH the two homomorphisms:
p_{1}(g,h) = g
p_{2}(g,h) = h
are very important: these are called PROJECTIONS.
it turns out that these projections actually DEFINE GxH:
suppose we have a third group K, with two group homomphisms φ_{1}: K-->G, and φ_{2}:K-->H.
i claim there is a UNIQUE homomorphism ψ:K-->GxH with:
p_{1}(ψ(k)) = φ_{1}(k) and
p_{2}(ψ(k)) = φ_{2}(k), for all k in K.
how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ_{1}(k),φ_{2}(k)) (convince yourself that this works, and IS a homomorphism).
in fact the properties:
p_{1}∘ψ = φ_{1}
p_{2}∘ψ = φ_{2}
force this definition on us. just as GxH is called the (direct) PRODUCT of G and H, we can call ψ the product of the homomorphisms φ_{1} and φ_{2}: ψ = φ_{1} x φ_{2}.
this is very similar to (in some ways "just the same") the projection of a point in the plane (x,y) onto either the x-axis, or the y-axis. in fact, it turns out that as (additive/abelian) groups, we have:
R^{2} ≅ R x R. the analogue of G* is in this case, the x-axis, and perhaps it should be clearer to you that the x-axis (the points in the plane (x,0)) is indeed isomorphic to the real line.
Sorry Deveno I've a question.
You said:
Is the choice of the function here arbitrary?
How do one choose a function if they suspect that two groups may be isomorphic? Is there any rule for that?
I've this question because I for this instance had difficulty figuring out the isomorphic function .
So my question: is there any rule that I need to follow for assigning one group's element to the other?
or is it always straightforward like you demonstrated in your answer to my question like this:
?
Are all the isomorphic functions this simple?
I found the answer. Not all isomorphic functions are this simple.
For example, the group of positive real numbers is isomorphic to the group of real numbers
The isomorphic function is
So all isomorphic functions are not simple and that's the answer to my question. You've to use trial and error to find the isomorphic function.
yes, and no.
proving two groups are isomorphic is not always as easy as it sounds. often, given two groups G and G', it's easier to show they AREN'T isomorphic, then it is to find an isomorphism.
some practical things to check first:
are the two groups of the same size (if one is infinite and one is finite, that settles it)?
are both groups cyclic, or non-cyclic?
are both groups abelian, or non-abelian? (if they are both abelian and finitely-generated, there is a structure theorem that allows us to tell if they are isomorphic or not).
the "bad case" for many practical purposes is two finite non-abelian groups of the same order: we may have two "presentations" of the same group (that is the group may be defined on a generating set with relations), and it can be darned difficult to TELL.
in THIS particular case, the isomorphism of G* = Gx{e_{H}} with G is pretty easy to figure out. such is usually NOT the case, and is one of the reasons a direct product is a "nice" way to make a bigger group from two smaller ones (we can recover the "factors" fairly easily). so if we know a group K = GxH, where G and H are groups we understand well, we've got a good grip on what K is like.
Thank you Deveno for your informational post. I understand what you mean. Sorry for late reply. I edit my last post here. I told to use "trial and error" to find isomorphism. After reading your(Deveno) post I think my "trial and error" method is wrong. Thanks for correcting my concept about this.