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Homework Statement
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?
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.Why are you re-posting this problem by starting a new thread?
https://www.physicsforums.com/showthread.php?t=543779"
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:Hi jdinatale!
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} = a^{n}/b^{n}.
What is the inverse of 0?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:
http://i3.photobucket.com/albums/y89/jdinatale/dookie.jpg" [Broken]
Yes, that is equivalent.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.
Uuuh, find another professor??? Seriously...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:
What is the inverse of 0?
ℂ* or ℂ^{x} is the set ℂ with all non-invertible elements removed.
Yes, that is equivalent.
Note that you can use (ab)^{n}=a^{n}b^{n} without proving it, since it is a property of ℂ.
Same thing for (a^{-1})^{n}=1/a^{n} as long as a≠0.
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!Uuuh, find another professor??? Seriously...
Your "group" does not have an inverse. There is no complex number x such that 0x=1.
If a is in G, then a^{n} - 1 = 0, so a^{n} = 1 and a ≠ 0.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/a^{n} = a^{n - 1}. My difficulty lies in proving that a^{n - 1} is an element of G.