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For the isomorphism of $U_8$ (where $U_8 = \{ z \in \mathbb{C} | z^8 = 1 \}$) with $\mathbb{Z}/8\mathbb Z$ in which $\zeta =e^{i2\pi/8} \mapsto 5$ and $\zeta \cdot \zeta = 5 +_8 5 =2$

Why is $\zeta^0 = 1$?

I cannot figure out how we get 0 in this.

EDIT: Directly quoting the problem

It can be shown that there is an isomprhism of $U_8$ with $\mathbb{Z}_8$ in which $\zeta = e^{i\pi/4} \leftrightarrow 5$ and $\zeta^2 = 2$. For $m=0$, we have $\zeta^0 = 1$ and for $m = 3$ we have $\zeta^2 \cdot \zeta =2 +_8 5=7$ and similarly $\zeta^4 = \zeta^2 \zeta^2 = 2 +_8 2 = 4$


Why is $2+_8 5= 7$? Why isn't it $2 + 5 - 8 = -1 $?

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$U_8$ is a finite group of order 8 while $\Bbb{Z}$ is not finite and hence the two cannot be isomorphic. – user38268 Sep 20 '12 at 5:41
I take it your isomorphism is actually with ${\bf Z}/8{\bf Z}$, not with $\bf Z$. But $\zeta^0=1$ because $z^0=1$ for all non-zero complex numbers $z$. – Gerry Myerson Sep 20 '12 at 5:42
I should add to the question that I have no knowledge of what a "group" is yet. And I pulled this stuff out of a section under "Roots of Unity" (so complex analysis) – Hawk Sep 20 '12 at 5:42
If the problem really says $\zeta^2=2$, then buy yourself another book, because that's nonsense. Anyway, I've told you, a couple of comments up, why $\zeta^0=1$, and it has nothing to do with the isomorphism. Are you OK with that reason? – Gerry Myerson Sep 20 '12 at 13:23
up vote 1 down vote accepted

For your additional question: (2 + 5) mod 8 = 7 mod 8, but 7 and -1 are in the same conjugacy class.

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But is there reason why we favor 7 over -1? – Hawk Sep 20 '12 at 6:47
We usually define the integers modulo 8 by {0,1,...,7} so the author is just sticking to standard notation. That's just what I think. – user41442 Sep 20 '12 at 6:53
We usually prefer $-1$. – Berci Oct 24 '12 at 18:11

In a homomorphism $f:G\to H$ of groups you always have $f(x^n)=f(x)^n$ for all $x\in G$, $n\in \mathbb Z$. Since here $H=\mathbb Z/8\mathbb Z$ is written additively, the power is in fact multplication, i.e. $f(x^n)=n\cdot f(x)$ for $x\in G$, $n\in \mathbb Z$. Especially, $f(\zeta^0)=0\cdot f(\zeta)=0$.

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Never came across the word the term "homomorphism" yet. – Hawk Sep 20 '12 at 6:00

$\zeta\cdot\zeta^{-1}=\zeta^{1-1}=\zeta^0=1$ since 1 is the identity in $U_8$, although I feel like this is missing part of your question. Could you rephrase it possibly?

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I'll quote directly from the book. It's an example done. I just don't see where the 0 or -1 is defined in the isomorphism. – Hawk Sep 20 '12 at 5:49

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