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

[jdinatale]

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?

 

[SammyS]

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

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

 

[jdinatale]

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.

 

[I like Serena]

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.

 

[jdinatale]

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]

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"

What is the inverse of 0?
ℂ* or ℂ^x 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.

 

[micromass]

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.

 

[jdinatale]

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=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 = a^{n - 1}. My difficulty lies in proving that a^{n - 1} is an element of G.

 

[I like Serena]

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 = a^{n - 1}. My difficulty lies in proving that a^{n - 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 a^n-1:
(a^n-1)ⁿ - 1 = (aⁿ)^n-1 - 1 = 1^n-1 - 1 = 0