Abelian group - set $G$ of all complex roots of $1$, of all possible orders

Question

Prove that the set G of all complex roots of $1$, of all possible orders is an abelian group with respect to multiplication.

So I wrote conditions to be an abelian group

1) $\forall a, b \in G, a*b \in G$

2) $\forall a,b,c \in G, (a*b)*c = a* (b*c)$

3) $\exists e, \forall a \in G, a*e=e*a=a$

4) $\exists b \forall a \in G, a*b = b*a=e$.

5) $\forall a, b \in G, a*b = b * a$.

I got the 1st and 2nd one. By the fact that $a=1^{1/n}*1^{1/m}=1^{1/n+1/m}$. So that still is a root of $1$ of some order $n+m$. Is that correct?

I stuck on the 3rd one with identity element. I was thinking about $e=1$ but $1^{1/n}*1=1^{1/n+1}$

Any tips?

// I think that I have realized that my approach to use $1^{1/n}$ instead of some $z^{n}=1$ might be incorrect.

Answer 1

No, your justification is not correct. If $a^m=b^n=1$, then $(ab)^{mn}=(a^m)^n(b^n)^m=1$.

The identity element will be $1$, of course.

It must bechecked too that $a\in G\implies a^{-1}\in G$. But this is easy: $$a^n=1\implies(a^{-1})^n=\left(\frac1a\right)^n=\frac1{a^n}=\frac11=1.$$

Answer 2

Since $1^{1/n}\times 1^{1/m}=1^{(n+m)/nm}$, what you can really say is that the order of that element divides $nm$.

In your calculation with the identity, you've made a mistake. The number $1$ itself should be thought of as $1^0$, so it works as an identity. It is, after all, the familiar number $1$.

Answer 3

Guide:

Try to use the following charactherization of complex root of $1$:

• Complex root of $1$ can be written as $\exp\left( j\frac{2\pi i}{n}\right)$ where $j \in \mathbb{Z}$.

For example to prove the $1$st property.

$$\exp\left( j_1\frac{2\pi i}{n}\right)\exp\left( j_2\frac{2\pi i}{n}\right)=\exp\left(( j_1+j_2)\frac{2\pi i}{n}\right)$$

since $j_1+j_2 \in \mathbb{Z}$, it is again a root of unity.