# I'm having difficulty proving this is a group under multiplication

## 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?

Why are you re-posting this problem by starting a new thread?

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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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 = an/bn.

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 = an/bn.

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.

I like Serena
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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]

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

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.

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.

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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:

Uuuh, find another professor??? Seriously...

Your "group" does not have an inverse. There is no complex number x such that 0x=1.

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.

Uuuh, find another professor??? Seriously...

Your "group" does not have an inverse. There is no complex number x such that 0x=1.

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.

I like Serena
Homework Helper
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.

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

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