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Homework Help: I'm having difficulty proving this is a group under multiplication

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    3. The attempt at a solution

    The above is my work. The problem is trying to make sure the inverse is element of the subgroup.

    Any leads? Also, was I correct in solving for x and y in that fashion?
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  3. Oct 25, 2011 #2


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    Last edited by a moderator: Apr 26, 2017
  4. Oct 25, 2011 #3
    The two images got mixed up with photobucket (they had the same file name, so they showed up in both threads), but they were two different problems. I just fixed it.
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  5. Oct 25, 2011 #4

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    Hi jdinatale! :smile:

    I'm afraid ℂ is not a group under multiplication.
    Perhaps you meant ℂ*?

    Then you're referring to the Subgroup Test.

    I quote wiki: "Let G be a group and let H be a nonempty subset of G. If for all a and b in H, ab-1 is in H, then H is a subgroup of G."

    This does not appear to what you have done.

    Since we're talking ℂ*, you can use (ab-1)n = an/bn.
  6. Oct 25, 2011 #5
    I'm not sure what the notation ℂ* means, but the complex numbers ARE a group under multiplication. I even wrote a proof of it. Plus my professor verified it:


    As far as the subgroup test - my book presents a logically equivalent version of the subgroup test. It states, paraphrased that if a subset H of a group G meets the following criteria, then H is a subgroup of G
    i) H is nonempty
    ii) for all a, b in H, ab is in H
    iii) for all a in H, a^-1 is in H

    Which is equivalent to the subgroup test on wikipedia.
  7. Oct 25, 2011 #6

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    What is the inverse of 0?
    ℂ* or ℂx is the set ℂ with all non-invertible elements removed.

    Yes, that is equivalent.

    Note that you can use (ab)n=anbn without proving it, since it is a property of ℂ.
    Same thing for (a-1)n=1/an as long as a≠0.
    Last edited by a moderator: May 5, 2017
  8. Oct 25, 2011 #7
    Uuuh, find another professor??? Seriously...

    Your "group" does not have an inverse. There is no complex number x such that 0x=1.
  9. Oct 25, 2011 #8
    Thanks, you guys are absolutely right! Seriously, my professor told me that the complex numbers were a group under multiplication, and that I didn't even need to prove it as a lemma when writing proofs for next time. I turned in the very lemma I posted above, and he didn't even mark off!

    But anyways, back to the original problem. (a-1)n=1/an = an - 1. My difficulty lies in proving that an - 1 is an element of G.
  10. Oct 25, 2011 #9

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    If a is in G, then an - 1 = 0, so an = 1 and a ≠ 0.

    (a-1)n - 1 = 1/an - 1 = 1/1 - 1 = 0

    So a-1 is also in G.

    EDIT: As for an-1:
    (an-1)n - 1 = (an)n-1 - 1 = 1n-1 - 1 = 0
    Last edited: Oct 25, 2011
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