# The group U_n

1. Feb 11, 2005

### hedlund

Prove / disprove that $$\left< U_n, \cdot \right>$$ is a group. The elements of $$U_n$$ is the solutions to $$x^n = 1$$.

Example:
$$\left< U_4, \cdot \right>$$ is the solutions to $$x^4 = 1$$, $$U_4 = \left\{ 1, -1, i, -i \right\}$$. And here $$\cdot$$ is multiplication. So I'm wondering if this is enough to prove that $$\left< U_n, \cdot \right>$$ is a group ...

1. There exists $$e \in U_n$$ such that ae=ea=a for all a. This can be shown to be e=1 since $$1\cdot a = a \cdot 1 = a$$. We know that for all n then 1^n = 1 ... hence $$1 \in U_n$$

2. Closure, if $$a,b, \in U_n$$ then $$a \cdot b \in U_n$$. This must be true since if $$a^n = b^n = 1$$ then $$\left( a \cdot b \right)^n = a^n \cdot b^n = 1$$ hence it is closed under multiplication

3. Existence of inverse for all elements, this must be true since if $$a^n = 1$$ then we know from the fact that the elements of $$U_n$$ is the ones of the form x^n = 1. So all x satisfying this must have |x| = 1. Hence if $$a^n = 1$$ for $$a = e^{iv}$$ for some v then $$a' = e^{-iv}$$ and this is the conjugate of a. Ie $$a' = \bar{a}$$. It can be prove that if a is a root of a polynom with real coefficients then $$\bar{a}$$ must also be a solution. Hence a' exists.

4. Associative, this is true due to that normal multiplication is assocative

The one that I'm not sure about is 2, the one about closeure ... but I don't know.

2. Feb 11, 2005

### Hurkyl

Staff Emeritus

3. Feb 11, 2005

### matt grime

And proving for n=4 doesn't prove for all n, obviously. However it should be clear to you that you don't actually use the "fourness" of 4 in the proof.
The proof of inverses is unnecessarily long: if x^n = 1, then x^{-n}=1 too. hence it is closed under inverses. And, as Hurkyl says: what is wrong with 2, it looks fine to me.

4. Feb 12, 2005

### hedlund

Yeah your proof is easier for closure of inverses ... #2 just didn't feel right, just a feeling ... it seems rights but feels wrong. Hard to explain

5. Feb 14, 2005

### lokisapocalypse

And if all else fails for closure, you could list all the combinations of the elements and show that they each give you something back in Un. Although that would be the REALLY long way to go.

6. Feb 14, 2005

### matt grime

Well, it is correct. You take two numbers whose n'th power is 1 and show their product's n'th power is 1. Thus it satisfies all the criteria for being in the group.

7. Feb 14, 2005

### chronon

Maybe #2 didn't feel right because it relied on the commutativity of the underlying group (the complex numbers), and so wouldn't generalize to an arbitrary underlying group, whereas the other items would

8. Feb 14, 2005

### matt grime

In my experience people doing group questions at this level often fail to notice that there are such things as noncommutative groups.

9. Feb 14, 2005

### AKG

If x is in your group, then it's inverse 1/x is in the group since (1/x)^n = 1^n/x^n = 1/1 = 1.