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I am currently studying abstract algebra through Evan Chen's Napkin. He says under example 1.1.9 that:

Let $p$ be a prime. Consider the non-zero residues modulo $p$, which we denote by $(\mathbb{Z}/p\mathbb{Z})^{\times}$. Then $((\mathbb{Z}/p\mathbb{Z})^{\times},\times)$ is a group.

It then asks under question 1.1.10:

Why do we need the fact that $p$ is prime?

which, to my understanding, is an exercise to the reader. Since the binary operation is multiplication, I am pretty sure we are talking about modular multiplication, which looks like:

enter image description here

The above is modular multiplication modulo 10. What I am getting confused about however is the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$. In the book, it is defined as the non-zero residues modulo $p$, i.e.:

$$(\mathbb{Z}/p\mathbb{Z})^{\times}=(1,2,3,...,n-1)$$

However, on Wikipedia, the multiplicative group of integers modulo $n$ is the set of numbers from $(1,2,3,...,n-1)$ coprime to $n$. Assuming it is the book's definition, my proof would be:

If $n$ is a positive integer not prime, then there exist 2 numbers $a$ and $b$ such that $a\times{b}=n$. However, since neither $n$ or 0 is found in the set $(1,2,3,...,n-1)$, then closure of the binary operation, multiplication, is not achieved. Furthermore, since the identity is 1, then there doesn't exist an inverse for all numbers since there is at least one factor of n in $(1,2,3,...,n-1)$, which we will denote $d$, as $d\times{k}\mod(n)\neq{1}$ for any positive integer $k$. $(\mathbb{Z}/n\mathbb{Z})^{\times},\times)$is therefore not a group.

Instead, choose a positive integer $p$. Then closure is achieved since two numbers $a$ and $b$ can never multiply to $p$ as that would violate the statement that $p$ is prime.

I am unable to continue the proof from here since I am unsure how to prove an inverse in $(1,2,3,...,p-1)$ exists for all integers in $(1,2,3,...,p-1)$. It is obvious that the identity is 1 and there exists associativity since the operation is multiplication though for both cases. To summarise, when answering could you please:

  • Clarify the concrete meaning of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ by writing out the set.
  • Check that the binary operation within this group is, in fact, modular multiplication.
  • Check whether the first bit of my proof is valid and give hints to finishing the proof or provide an alternative proof (though my knowledge in abstract algebra is a bit limited considering I just started learning it).

I apologise if I have a silly misconception or one that is quite trivial, though I have started looking at cyclic groups and the book has not yet introduced me to quotient groups. I thank anyone in advance for their help!

Update following suggestion Yourong 'DZR' Zang

To prove the statement suggested, we define the sets:

$$[n]=\{n+pk_1:k_1\in\mathbb(Z)\}$$ $$[m]=\{m+pk_2:k_2\in\mathbb(Z)\}$$ $$[a]=\{nm+pk_3:k_3\in\mathbb(Z)\}$$

Obviously $[a]\equiv[n][m]\mod{p}$ since $(nm+pk_3)\mod{p}=nm$ and $((n+pk_1)(m+pk_2))\mod{p}=(nm+nk_2p+mk_1p+k_1k_2p^2)\mod{p})=nm$. We now prove the sets are equal. Since $(n+pk_1)(m+pk_2)=(nm+nk_2p+mk_1p+k_1k_2p^2)=nm+p(nk_2+mk_1+k_1k_2p)$, letting $k_3=nk_2+mk_1+k_1k_2p$ lets the result directly follow.

Now I have proven this statement, I am having trouble identifying how this proves that there exists an inverse for all numbers in $(\mathbb{Z}/p\mathbb{Z})^{\times}$ since that would require the existence of a number $h\in(\mathbb{Z}/p\mathbb{Z})^{\times}$ such that $nh\equiv{1}\mod{p},\forall{n}\in(\mathbb{Z}/p\mathbb{Z})^{\times}$.

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    $\begingroup$ Apply Bezout's identity. Think about $\gcd(p,n)$ where $n<p$. The set $(\mathbb{Z}/p\mathbb{Z})^\times$ is the multiplicative subset of the quotient $\mathbb{Z}$ mod $p\mathbb{Z}$. It contains equivalence classes $\{\bar{1},\dots,\overline{p-1}\}$. Each equivalence class represents a set $\bar{n}=\{n+pk:k\in\mathbb{Z}\}$. You could prove that $\bar{n}\bar{m}=\overline{nm \bmod{p} }$. $\endgroup$
    – user672528
    Commented Jul 7, 2020 at 0:37
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    $\begingroup$ The notation $\times$ is actually designated to indicate that this set contains the multiplicative inverse of each of its elements. $\endgroup$
    – user672528
    Commented Jul 7, 2020 at 0:38
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    $\begingroup$ About the part you mentioned from Wikipedia : "the multiplicative group of integers modulo $n$ is the set of numbers from $(1,2,3,\cdots,n−1)$ coprime to $n$", the multiplicative group modulo $n$, is not the actual set $\{1,2,\cdots,n-1=[n-1]\}$ (that's just a notation), it is only a subset of this set, a subset containing only the integers in $[n-1]$ that are coprime to $n$. Take that into consideration, then your proof for $\left( \left( \mathbb{Z}/n\mathbb{Z}\right)^{\times}, \times \right)$ should change. Check en.wikipedia.org/wiki/B%C3%A9zout%27s_identity & update the question :) $\endgroup$ Commented Jul 7, 2020 at 0:39
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    $\begingroup$ Yes that is correct. We usually write $\bar{n}$ for the equivalence class with residue $n$. Some people also write $[n]$ or just $n$ for convenience. $\endgroup$
    – user672528
    Commented Jul 7, 2020 at 0:45
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    $\begingroup$ I apologize for that. I just want to say their product is $\bar{n}\bar{m}=\overline{a}$ where $a\equiv mn\pmod{p}$. The representative doesn't matter in this case. The notation I used was not a good one. Just ignore it. $\endgroup$
    – user672528
    Commented Jul 7, 2020 at 0:52

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To answer your question why you need $p$ to be a prime. If $p$ is a prime then $\mathbb{Z}/p\mathbb{Z}$ is a field. Let $\mathbb{F}=\mathbb{Z}/p\mathbb{Z}.$ If you look up the definition of a field, it if $\mathbb{F}$ is a field, then $\mathbb{F}^{\times}$ is a group under mutliplication.

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