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Basic proof for Homomorphism of Abelian Groups

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f : G → H be a homomorphism of Abelian groups.

    1. Show that f (0) = 0.
    2. Show that f (−x) = −f (x) for each x ∈ G.

    2. Relevant equations

    3. The attempt at a solution

    My background in topology / group theory is next to nothing.

    1. Show that f(0) = 0. My attempt is as follows:

    f(x+y) = f(x) + f(y)

    Let y = 0

    f(x+0) = f(x) + f(0)
    f(x) -f(x) = f(x) -f(x) + f(0)
    0 = 0 + f(0)
    0 = f(0)

    Is this a valid proof?

    2. I tried the following:

    f(-x) = f(-1 * x) = f(-1)*f(x) = -1 * f(x) = -f(x)

    I'm fairly certain that for a homomorphism, f(1) = 1. Is that correct? I'm guessing that f(-1) = -1, though I'm not sure whether this is correct.

    Can somebody please provide me with some comments or suggestions?

    Thank you!
     
    Last edited: Jan 6, 2009
  2. jcsd
  3. Jan 6, 2009 #2
    Well f(n) = n right? Since let's take f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2.

    You can check that f(1) = 1 since f(1) = f(1*1) = f(1)*f(1) therefore 1 = f(1)

    I guess I'm not remembering my algebra right what's the operation for G and H. In one case we seem to be dealing with addition, the other multiplication or does it not matter?
     
    Last edited: Jan 6, 2009
  4. Jan 6, 2009 #3
    I'm not sure that f(n) = n. That isn't given in the problem statement.

    Verifying that f(1) = 1 isn't difficult, but I do appreciate your proof.

    For a homomorphism, f(x+y) = f(x) + f(y) because f maps G → H

    How do I show that f(-x) = -f(x)?

    For that matter, is it true that f(-1) = -1?

    Thank you!
     
    Last edited: Jan 6, 2009
  5. Jan 6, 2009 #4

    Hurkyl

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    This doesn't make sense. What is -1? What is *? In the problem at hand, there is no such thing as '1'.
     
  6. Jan 6, 2009 #5
    Hurkyl,

    I am at the beginning of learning group theory on my own, so my questions may seem a bit basic.

    Why doesn't f(-1 * x) make sense? Does f(-x) actually mean the f(inverse of x)?

    With regards to '*' and '-1', I meant to factor '-x' into two separate parts so that I can somehow apply the axiom f(x+y) = f(x) + f(y). Based upon your reply, I believe that you're telling me that '*' and '-1' are not valid operations in this problem?

    Why is there no such thing as '1'? I thought that '1' was the identity element.

    Thank you for your help.
     
    Last edited: Jan 6, 2009
  7. Jan 6, 2009 #6
    I thought I proved that f(n) = n, it doesn't need to be given since

    [tex] f(n) = f(\sum_{i=1}^{n}{1}) = \sum_{i=1}^{n}{f(1)} = \sum_{i=1}^{n}{1} = n [/tex]

    1 doesn't make sense because 1 is the identity element for multiplication. For hom f(e) = e. So if you have addition then e is 0 hence f(0) = 0.

    That's why I asked which operation are you working under.
     
  8. Jan 6, 2009 #7

    radou

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    First of all, if you have chosen additive notation for an Abelian group, you should be perfectly consistent with it.

    OK, for the second one, what does f(-x) + f(x) equal? (Hint: f is a group homomorphism.)
     
  9. Jan 6, 2009 #8

    Hurkyl

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    The name of the identity element is whatever name you assign to it. You were previously calling the identity element '0'. Furthermore, you had been using '+' for the binary operation, and '-' for the inverse operation.

    If you really meant to use '1' for the identity, and '*' for the binary operation, then nothing you wrote previously makes sense. (And the way you used '-' would be nonsensical)


    There are structures called 'rings' that have two binary operations (usually written '+' and '*'), and the corresponding two identities (usually written '0' and '1'), and given the identities for rings, mixing them like that would make sense. But the question isn't about rings....
     
  10. Jan 6, 2009 #9
    Ok, so I am working under the operation of addition because I'm given that f(x+y) = f(x) + f(y) for f: G-->H.

    Thanks to your clarification, I now understand the difference in identity elements between the additional and multiplication operations.

    Now, I'm working with addition and am trying to show that f(-x) = -f(x).

    Does -x mean the inverse of x?

    f(-x) = -f(x)
    = 0 - f(x)
    = f(0) - f(x)

    But where do I go from there?
     
  11. Jan 6, 2009 #10
    Wait why are you starting with what you are trying to prove? If you are doing a direct proof, start with f(-x) and through a series of logical connections, get to -f(x). Use the hint given above, those people seem to know more than I do :)

    And yes, -x is the additive inverse of x.

    So perhaps 0 = f(0) = f(x - x) = f(x) + f(-x) therefore f(x) = -f(x)
     
  12. Jan 6, 2009 #11
    Ok, I've heard of rings and your answer makes sense.

    I believe that I've proved f(0) = 0 as follows:

    f(x + y) = f(x) + f(y)

    Take y = 0
    f(x + 0) = f(x) + f(0)
    f(x) = f(x) + f(0)
    f(x) - f(x) = f(x) - f(x) + f(0)
    0 = f(0), which is what I was trying to prove.

    But to prove -f(x) = f(-x), I'm at a loss.

    Hurkyl helped tremendously with this:

    0 = f(0)
    0 = f(x - x)
    0 = f(x) + f(-x)
    -f(x) = -f(x) + f(x) + f(-x)
    -f(x) = 0 + f(-x)
    -f(x) = f(-x), which is what I was trying to prove.

    Do both of these look valid?

    Thank you!
     
    Last edited: Jan 6, 2009
  13. Jan 6, 2009 #12
    Thank you, your proof seems so simple.

    I honestly appreciate your help and feel humbled. :)

    Regarding your question as to why I started with what I was trying to prove: The answer is that I'm new to proofs and am trying to learn how to do them. I've been out of school for about 4 years and took math up to linear algebra. I didn't have to do any proof based math in the pursuit of my degree.
     
  14. Jan 6, 2009 #13
    That's an excellent hint and I wish that I would have seen it a little earlier. I appreciate your help!
     
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