What exactly is the group of invertible residues modulo n under multiplication? It was brought up in lecture, and I looked for it online but could not make sense of the explanations. Can anyone help me explain this concept? Thanks!
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By the Bezout identity $\mathbb{Z}n^\times$ is the set of integers $k \in {1, \ldots n}$ such that $gcd(n,k)= 1$. Because of the chinese remainder theorem it is isomorphic to the direct product $\mathbb{Z}{p_1^{e_1}}^\times \ldots \times \mathbb{Z}_{p_m^{e_m}}^\times$ where $n = \prod_i p_i^{e_i}$ is its prime factorization – reuns Feb 15 '17 at 01:17
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This is a good example for looking up wikipedia article for the multiplicative group of integers modulo $n$. It has also many examples. This group has also beed studied intensively on MSE, e.g., here, or here, or here. The notation often is $U_n$, the unit group of the ring $\mathbb{Z}/n$.
Dietrich Burde
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The group of all numbers that have an inverse under mod n multiplication.
For example, the inverse of $2 (mod 5) = 3$, because $2*3 = 1 (mod 5)$. Therefore 2 has an inverse mod 5.
It turns out that x has an inverse mod n, if and only if x and n are coprime.
Kaynex
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I think that's what they meant: $$(\mathbb Z / p\mathbb Z)^*=\{\bar n\in \mathbb Z/p\mathbb Z: \exists \bar m \in \mathbb Z/p\mathbb Z: \bar n\cdot \bar m=1\}$$
A. Salguero-Alarcón
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Let me try to put into plain English to make sure I got what you mean. All elements within this group are coprime? – richcao Feb 14 '17 at 20:04
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That might be plain English, but it does not make much sense. Two integers can be coprime, but it makes no sense to say that a single group element is coprime. Coprime to what? – Derek Holt Feb 15 '17 at 00:53
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