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Just as the title says, I'm asked to prove the following:

For $\alpha\in\mathbb{T}$, where $\mathbb{T}=\{z\in\mathbb{C}:|z|=1\}$, prove that $\alpha^n=\alpha^m=1$, only if $\alpha^d=1$, where $d=\gcd{(m,n)}$.

Here is my attempt at the proof.

If $d=\gcd{(m,n)}$, then we can write $d$, as $d=rm+sn$, for some $r,s\in\mathbb{Z}$. Hence we can write $\alpha^d=\alpha^{rm+sn}=\alpha^{rm}\alpha^{sn}$. Then, using the fact that $|\alpha|=1$, because $\alpha\in\mathbb{T}$, we can let $\alpha=e^{ix}$. Then we have that $\alpha^{rm}\alpha^{sn}=e^{ix(rm+ns)}=e^{irmx}e^{isnx}=1$.

At this point, I'm not sure exactly how to proceed, or even if I am ont the correct track. Any help would be appreciated.

quanticbolt
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  • No need to use the exponential form: just notice that $\alpha^{rm}=(\alpha^m)^r=1^r$, and similarly with $\alpha^{sn}$. – Leo163 Oct 01 '17 at 19:15

2 Answers2

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This is just a property of cyclic groups, as pointed out in the comments. Then if $m = du, n = dv$ we have $\alpha^{m} = (\alpha^d)^u = 1^u = 1$, and for the other implication, namely if $\alpha^m = \alpha^n = 1$, use Bezout's lemma as you did.

StefanH
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You're close. As Leo said in his comment, you don't need $\alpha = e^{ix}$; just use $d = rm + sn$:

$\alpha^d = \alpha^{rm + sn} = \alpha^{rm} \alpha^{sn} = (\alpha^m)^r (\alpha^n)^s = 1^r \cdot 1^s = 1. \tag 1$

That $\alpha^d = 1 \Longrightarrow \alpha^n = \alpha^m = 1$ is almost self-evident, since $d \mid m$ and $d\mid n$.

Robert Lewis
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  • I understand how to prove this. but what is the intuition for this to be true? can you explain the intuition please – user20194358 Mar 06 '23 at 22:10