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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?

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- Thread starter jdinatale
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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

SammyS

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

https://www.physicsforums.com/showthread.php?t=543779"

https://www.physicsforums.com/showthread.php?t=543779"

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

https://www.physicsforums.com/showthread.php?t=543779"

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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- #4

I like Serena

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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

This does not appear to what you have done.

Since we're talking ℂ*, you can use (ab

- #5

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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}.

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.

- #6

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 ℂ

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)

Same thing for (a

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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.

- #8

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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.

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

- #9

I like Serena

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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.

If a is in G, then a

(a

So a

EDIT: As for a

(a

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