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I am trying to learn some algebra to prepare myself for graduate school.

I am doing the problem from The fundamental Theorem of Algebra by Benjamin Fine & Gerhard Rosenberger. Don't worry I am not doing homework.

Problem 2.5: Let $w^3=1$, $w \neq1$. Show then that $1+w+w^2=0$.

That was very easy.
My question is about $w^3=1,w\neq1$ does that map to $C_3$.

because w would generate the group? by the fact that its not the identity?

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    We are not worried with people doing their homework _on their own_; we are only worried about their posting it here and getting us to write out full solution showing all steps. In fact, we will be glad if people did their homeworks!2012-01-31

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Yes, your understanding is correct. Indeed, for any $n$, the set of $\zeta \in \mathbb{C}$ satisfying $\zeta^n=1$ form a cyclic group of order $n$ called the group of $n^{th}$ roots of unity. But we can't pick any $\zeta \ne 1$ as a generator in general; for example in the case $n = 4$, $-1$ certainly satisfies $(-1)^4 = 1$, but it only generates a group of order 2. A generator of one of these groups goes by the term "primitive $n^{th}$ root of unity".

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    @FortuonPaendrag thanks, I thought so, but I am don't know the accepted usages of all the Greek letters, yet.2012-01-31