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I read the following definition in a notebook.

The set $\mathbb{Z}^{*}_{n}$ that consists of all integers $i = 0, 1, ..., n-1$ for which the $\operatorname{gcd}(i, n) = 1$ forms an abelian group under the binary operation multiplicaiton modulo $n$. The identity element, of course, is $1$.

But I think it should be,

The set $\mathbb{Z}^{*}_{n}$ that consists of all integers $i = 1, ..., n-1$ for which the $\operatorname{gcd}(i, n) = 1$ forms an abelian group under the binary operation multiplicaiton modulo $n$. The identity element, of course, is $1$.

Or could it be that both the definition are valid?

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    $\begingroup$ Yes, you're right. $0$ wouldn't have an inverse. $\endgroup$
    – saulspatz
    Jan 31, 2021 at 13:07

1 Answer 1

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Both are valid. Note that $\operatorname{gcd}(0,n)=n\ne1$ (for $n>1$; but this is not serious restriction). Hence $0$ is excluded anyways.

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  • $\begingroup$ Hmm, but if "0 is excluded anyways" (as you pointed out), then won't it be contradictory to say that the set consists of "all integers i=0, 1, ..., n-1" (as the textbook says)? $\endgroup$
    – DaveIdito
    Jan 31, 2021 at 13:11
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    $\begingroup$ @DaveIdito Careful; it consists of "all integers $i=0,1,...,n-1$ for which the $\operatorname{gcd}(i,n)=1$". Take $n=4$ and $i=2$ to see that's this does not include all these integers necessarily. $\endgroup$
    – mrtaurho
    Jan 31, 2021 at 13:13
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    $\begingroup$ @DaveIdito there's no difference if you add the given restriction "for which $\gcd(i,n)=1$" (provided $n>1$). For instance, with $n=8$, we do not talak about all the integers $i=1,2,3,4,5,6,7$ either, but only of all of these that are odd. $\endgroup$ Jan 31, 2021 at 13:13
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    $\begingroup$ No that's not contradictory. On the contrary you have to include $0$ here if you want the statement to be valid for $n=1$. $\endgroup$
    – Derek Holt
    Jan 31, 2021 at 13:14
  • $\begingroup$ Thanks. Got it! $\endgroup$
    – DaveIdito
    Jan 31, 2021 at 13:15

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