How binary operation defined as $\operatorname{gcd}(a,n)=1$ is same as multiplication modulo $n$?
I have recently started learning group theory on my own. There is a problem that the set of all numbers coprime to $n$ is a group.
To prove this,
$$(a,n)=1 = (b,n) \implies (ab,n) =1$$
If $(ab, n) = d > 1$ then there exists some prime $p$ that divides $d$ and so $p$ divides $a$ or $b$ and $p > 1$. This leads to contradiction and so closure is satisfied
$(1, n) = 1$ and so $1$ is identity element in the set of all numbers coprime to $n$.
But in my textbook while proving inverse, the author suddenly takes binary operation defined by $\operatorname{gcd}$ w.r.t. $n=1$ as multiplicative modulo $n$. How is it possible?
I have following doubts:
If we have a number "$a$" coprime to $n$ then how to show that there exists another number $b$ coprime to $n$ such that a (some binary operation defined by $\operatorname{gcd}$ w.r.t. $n$) $b=$ unity $=1$
How is this binary operation is nothing but multiplication modulo $n$
Kindly clarify.
Q2. Let $S=\{x:1\leq x<n,(x,n)=1\}$ and $\cdot$ is the multiplication modulo $n$. Then $(S,\cdot)$ form a group. Note that the elements in $S$ are the integers less than $n$ which are coprime to $n$. However, I think you misunderstood the operation defined by author. Please check again your reference book.
For example, take $n=8$. Then $S=\{1,3,5,7\}$. In $(S,\cdot),$ we have \begin{align}3\cdot5=7,\\ 3\cdot7=5,\\ 5\cdot7=3.\end{align} You can check that the operation is the usual multiplication modulo $8$.
Q1. Let $a\in S$. By how we define $S$, we have $(a,n)=1$. By Bezout Identity, there exist $x,y\in \Bbb{Z}$ such that $$ax+ny=1.$$ Let $b$ be the remainder of $x$ when divided by $n$.
Then we have $ab\equiv ax\equiv 1 (\bmod n)$. In other words, $b\in S$ such that that $a\cdot b=1$ in $(S,\cdot)$.