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Math Help - Why and how is G* isomorphic to G?

  1. #1
    Senior Member x3bnm's Avatar
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    Why and how is G* isomorphic to G?

    There is an example on pg-155-156 of Pinter's abstract algebra book which is:

    The Problem:

    Let G and H be any groups and consider their product G \times H.

    Remember that G \times H consists of all the ordered pairs (x,y) as x ranges over G and y ranges over H. You multiply ordered pairs by multiplying corresponding
    components; that is, the operation on G \times H is given by

    (a,b) \cdot (c,d) = (ac, bd)

    Now, let f be the function from G \times H onto H given by

    f(x, y) = y

    It is easy to check that f is a homomorphism. Furthermore, (x, y) is in the kernel of f if and only if f(x,y) = y = e. This means that the kernel of f consists of all the ordered pairs whose
    second component is e. Call this kernel G^{*}; then

    G^{*} = \{(x,e) : x \in G \}

    We symbolize all this by writing

    f: (G \times H) \xrightarrow[ G^{*}\,\,\,\,\,\,\,     ]{\,\,} H

    By the fundamental homomorphism theorem, we deduce that H \cong (G \times H)/G^{*}.

    [It is easy to see that G^{*} is an isomorphic copy of G; thus identifying G^{*} with G, we have shown that, roughly speaking, (G \times H)/G \cong H.]




    My question is: why and how is that G^{*} isomorphic to G?

    Can anyone kindly tell me: why's that G^{*} here isomorphic to G?
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  2. #2
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    Re: Why and how is G* isomorphic to G?

    You are honestly claiming that you don't see an isomorphism between \{(x,e_H)\mid x\in G\} and \{x\mid x\in G\} (here e_H is the unit of H)?
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by emakarov View Post
    You are honestly claiming that you don't see an isomorphism between \{(x,e_H)\mid x\in G\} and \{x\mid x\in G\} (here e_H is the unit of H)?
    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 G_1 and G_2 be groups. A bijective function f : G_1 \to G_2 with the property that for any two elements a and b in G_1,

    f(ab) = f(a)f(b)

    is called an isomorphism from G_1 to G_2.

    If there exists an isomorphism from G_1 to G_2, we say that G_1 is isomorphic to G_2.



    I can't find a way to show that f is both one-to-one and onto. I know that f(ab) = f(a)f(b) because of homomorphism.


    In other words how do I prove that:

    1) If f(a, e_1) = f(b, e_2) then (a, e_1) = (b, e_2) and
    2) There is a function y = f(x) for every y?
    Last edited by x3bnm; February 14th 2013 at 07:42 AM.
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  4. #4
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by x3bnm View Post
    I can't find a way to show that f is both one-to-one and onto.
    This depends on what f is. How do you define it?

    Quote Originally Posted by x3bnm View Post
    I know that f(ab) = f(a)f(b) because of homomorphism.
    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 G^*\ni(x,e)\mapsto x\in G as a potential isomorphism.
    Thanks from x3bnm
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  5. #5
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by emakarov View Post
    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 G^*\ni(x,e)\mapsto x\in G as a potential isomorphism.

    I think I'm causing confusion. Suppose there is an example like below:

    -----------------------------------------------------------------------------------------------------------------------------------
    Let G and H be any groups and consider their product G \times H.

    Remember that G \times H consists of all the ordered pairs (x,y) as x ranges over G and y ranges over H. You multiply ordered pairs by multiplying corresponding
    components; that is, the operation on G \times H is given by

    (a,b) \cdot (c,d) = (ac, bd)

    Now, let f be the function from G \times H onto H given by

    f(x, y) = y

    It is easy to check that f is a homomorphism. Furthermore, (x, y) is in the kernel of f if and only if f(x,y) = y = e. This means that the kernel of f consists of all the ordered pairs whose
    second component is e. Call this kernel G^{*}; then

    G^{*} = \{(x,e) : x \in G \}

    ...
    -----------------------------------------------------------------------------------------------------------------------------

    Now how do I mathematically prove that G^{*} is isomorphic to G? Is it possible that G^{*} is isomorphic to G?

    But the author(Pinter) said that on pg-156:

    "...... G^{*} is a isomorphic copy of G....." . Then where am I wrong?
    Last edited by x3bnm; February 14th 2013 at 09:23 AM.
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  6. #6
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by x3bnm View Post
    Now how do I mathematically prove that G^{*} is isomorphic to G?
    As I said,
    Quote Originally Posted by emakarov View Post
    I feel almost silly offering to consider the map G^*\ni(x,e)\mapsto x\in G as a potential isomorphism.
    Thanks from x3bnm
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  7. #7
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by emakarov View Post
    As I said,
    But the author Pinter on his book(pg-156) "A Book of Abstract Algebra" second edition said that " .......[tex]G^{*} is a isomorphic copy of G....".

    The word "isomorphic" and "isomorphic copy" are they same thing or two different things?
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  8. #8
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by x3bnm View Post
    The word "isomorphic" and "isomorphic copy" are they same thing or two different things?
    The same. The word "isomorphic" is a technical term, but "copy" is just a part of the natural language description.
    Thanks from x3bnm
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  9. #9
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    Re: Why and how is G* isomorphic to G?

    consider the mapping:

    h:G*--->G given by: h(g,eH) = 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,eH)*(g',eH)) = h((gg',eH)) = gg' = h((g,eH))h(g',eH))

    2) h is injective:

    suppose that h((g,eH)) = h((g',eH)). this means that g = g', so (g,eH) = (g',eH) (both coordinates are equal).

    3) h is surjective:

    suppose g is ANY element of G. then (g,eH) 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,eH). 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 eH, 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:

    p1(g,h) = g
    p2(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 φ1: K-->G, and φ2:K-->H.

    i claim there is a UNIQUE homomorphism ψ:K-->GxH with:

    p1(ψ(k)) = φ1(k) and
    p2(ψ(k)) = φ2(k), for all k in K.

    how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ1(k),φ2(k)) (convince yourself that this works, and IS a homomorphism).

    in fact the properties:

    p1∘ψ = φ1
    p2∘ψ = φ2

    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 φ1 and φ2: ψ = φ1 x φ2.

    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:

    R2 ≅ 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.
    Thanks from x3bnm
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  10. #10
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by Deveno View Post
    consider the mapping:

    h:G*--->G given by: h(g,eH) = 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,eH)*(g',eH)) = h((gg',eH)) = gg' = h((g,eH))h(g',eH))

    2) h is injective:

    suppose that h((g,eH)) = h((g',eH)). this means that g = g', so (g,eH) = (g',eH) (both coordinates are equal).

    3) h is surjective:

    suppose g is ANY element of G. then (g,eH) 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,eH). 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 eH, 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:

    p1(g,h) = g
    p2(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 φ1: K-->G, and φ2:K-->H.

    i claim there is a UNIQUE homomorphism ψ:K-->GxH with:

    p1(ψ(k)) = φ1(k) and
    p2(ψ(k)) = φ2(k), for all k in K.

    how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ1(k),φ2(k)) (convince yourself that this works, and IS a homomorphism).

    in fact the properties:

    p1∘ψ = φ1
    p2∘ψ = φ2

    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 φ1 and φ2: ψ = φ1 x φ2.

    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:

    R2 ≅ 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.
    Deveno, you must be a good teacher. How do you know I've problem with embeddings and projections? Thanks for taking the time explaining this and thanks for the proof. It is such a great help.
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  11. #11
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Sorry Deveno I've a question.

    You said:

    Quote Originally Posted by Deveno View Post

    consider the mapping:

    h:G*--->G given by: h(g,eH) = g.
    ......

    Is the choice of the function h here arbitrary?

    How do one choose a function h 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 h.

    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:

    h(g,e_H) = g?

    Are all the isomorphic functions this simple?
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  12. #12
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by x3bnm View Post
    Sorry Deveno I've a question.

    You said:




    Is the choice of the function h here arbitrary?

    How do one choose a function h 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 h.

    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:

    h(g,e_H) = g?

    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 \mathbf{R}^{+} is isomorphic to the group of real numbers \mathbf{R}

    The isomorphic function f is \log_{b}:\mathbf{R}^{+} \to \mathbf{R}

    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.
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  13. #13
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    Re: Why and how is G* isomorphic to G?

    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{eH} 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.
    Thanks from x3bnm
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  14. #14
    Senior Member x3bnm's Avatar
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    Re: Why and how is G* isomorphic to G?

    Quote Originally Posted by Deveno View Post
    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{eH} 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.
    Last edited by x3bnm; February 28th 2013 at 07:27 AM.
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