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I have to solve a problem which states that $(m,n) = 1$, but I have no idea what this means. Maybe, the problem itself will help:

Suppose $m, n \in \mathbb{Z}$, with $n > 0$ and $(m,n) = 1$. Show that:

$$\left \{ e^{2 \pi i m k/n}: 0 \leq k < n \right \} = \left \{ e^{2 \pi i j/n}: 0 \leq j < n\right \}$$

Thanks a lot for any help.

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It means that $\gcd(m,n) = 1$. –  JavaMan Mar 14 '11 at 15:41
    
@DJC: I would have voted for this answer! –  The Chaz 2.0 Mar 14 '11 at 15:47
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up vote 7 down vote accepted

The notation $\rm\ (a,b)\ $ can denote either $\rm\ gcd(a,b)\ $ or the ideal $\rm\ a\ \mathbb Z + b\ \mathbb Z\:.\ $ In $\rm\:\mathbb Z\:$ (or in any Bezout domain $\rm\:Z\:$) these denote essentially the same object since $\rm\ a\ Z + b\ Z\ =\ c\ Z\ \ \iff\ \ c = gcd(a,b)\:.$ The advantage of the notation is that it allows one to simultaneously prove results for both gcds and ideals by using only those laws that hold true for both. For example, see my post on the Freshman's dream $\rm\ (a,b)^n = (a^n,b^n)\ $ for gcds and invertible ideals. Such analogies are exploited to the hilt when one studies divisor theory.

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$(m,n)=1$ simply means that the G.C.D of $m,n$ is $1$. In other words they are relatively prime.

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