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As an abstract algebra exercise, I need to show that the $n$th roots of unity is indeed a subgroup of $\mathbb{C}^\times$.

The only part of the proof I am concerned about is the multiplicative inverse. Here is what I have stated so far.

Let $a$ be an $n$th root of unity. Then $a^n = aa^{n - 1} = 1$. We will show that $a^{n - 1}$ is an $n$th root of unity. Observe that $(a^{n - 1})^n = a^{n^2 - n} = a^{n^2}a^{-n}$.

I want to say that, since $a^{-n}$ is a complex number, $a^{-n} = \frac{1}{a^n}$. Then $a^{n^2}a^{-n} = (a^n)^n\frac{1}{a^n} = 1^n\frac{1}{1} = 1$. Hence $a^{n - 1}$ is an $n$th root of unity and every $n$th root of unity has a multiplicative inverse.

It seems to me that I am making assumptions about how exponents work in an abstract setting and I am wondering if that is OK to do so?

WLOG
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Joseph DiNatale
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  • If you're concerned about that, think of it this way: $a^{-n} = (a^n)^{-1}$. You can say this only because the $n$th roots of unity is an abelian group. – Lost Jun 07 '14 at 20:24
  • Lost - But wouldn't I first have to prove the $n$th roots of unity is an abelian group, which would be almost the exact same exercise as I am trying to show? – Joseph DiNatale Jun 07 '14 at 20:25
  • How about using polars, with $e^{i\theta}=1=e^{i2\pi}$? – user99680 Jun 07 '14 at 20:32
  • I guess @Lost meant that the whole group $\mathbb{C}^{\times}$ is an abelian group, not just the subgroup of the $n$th roots of unity. – Ulrik Jun 07 '14 at 20:47

2 Answers2

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You're using all sorts of facts about exponentiation in $\mathbb C$ that are true, but you're right that they need to be proven beforehand. Specifically, you're using these:

For any $n, m \in \mathbb Z$ and $x\in\mathbb C$: $$x^{n+m}=x^nx^m$$ $$(x^n)^m=x^{nm}$$

Note that $x^{-n}=1/x^n$ (where $n$ is positive) isn't true "because $x$ is a complex number", it's true because that's the definition of raising to a negative power. The fact that it's also true when $n$ is negative, however, is a theorem (that you aren't using here since your $n$ is positive).

You should prove the above facts yourself if you want to put your mind at ease. You'll need the fact that multiplication in $\mathbb C$ is associative and commutative, and you may find it easiest to separate all of them into different cases depending on the signs of $n$ and $m$. Oh, and $x^0=1$ is also just a definition, so you can assume that.

Jack M
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$a^{-1}$ is a $n$-th root of unit because $$(a^{-1})^n = (a^{-1})^n \cdot 1 = (a^{-1})^n \cdot a^n = (a^{-1}a)^n = 1^n = 1 $$

WLOG
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