1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
  2. jcsd
  3. Oct 25, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
    Last edited by a moderator: Apr 26, 2017
  5. Oct 25, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook