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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](https://en.wikipedia.org/wiki/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 – 2017-02-15

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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.

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I think that's what they meant: $$(\mathbb Z / p\mathbb Z)^*=\{\bar n\in \mathbb Z/p\mathbb Z: \exists \bar m \in \mathbb Z/p\mathbb Z: \bar n\cdot \bar m=1\}$$

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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? – 2017-02-14
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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? – 2017-02-15
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    Sorry, meant coprime relative to whatever p you set for $(Z/pZ)$ – 2017-02-15
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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 $\mathbb{Z}/n$.

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    I'll see if I can make sense of it. Thanks! – 2017-02-14
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    You are welcome. I think you can understand it well. And if something is not clear, you have a good reason to post another question. – 2017-02-14