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Lemma#3: If a is relatively prime to n, then multiplication by a permutes the elements 1,2,3,....,n-1.

I didn't understand what permutes means here. If I consider $\mathbb{Z_n}$ then any element a relatively prime to n has the same order as 1 and also it generates $\mathbb{Z_n}$. Thats the maximum I know. Please I need help to understand this.Thank you.

Bill Dubuque
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Sunit das
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    Take an example. Sat $n=5$ so we are looking at $(1,2,3,4)$. If we multiply by, say, $a=2$ ,we get $(2,4,6,8)$ which, $\pmod 5$ is $(2,4,1,3)$. Note that $(2,4,1,3)$ is a permutation of $(1,2,3,4)$. – lulu Nov 22 '20 at 21:18
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    A permutation of a set $S$ is a map $S\to S$ that is bijective. Note tha tfor finite sets, you need only check one of injective/surjective because bijective then follows. – Hagen von Eitzen Nov 22 '20 at 21:18
  • In this case I understand how the permutation is working. The question do I have is if I define a function $f_a$:{1,.....,n-1}$\rightarrow${1,2,....,n-1} s.t f(x)=a.x and show that this function is one-to-one will it be enough? – Sunit das Nov 22 '20 at 21:35
  • Yes! It is a very useful (and easy to prove) fact that, if $F$ is a function from one finite set to another then $F$ is one-to-one if and only if it is onto. Of course this is not true for infinite sets, but it is true for finite sets. here is a reference. – lulu Nov 22 '20 at 21:37
  • Yes I have done this in exercise and for finite sets. Thank you so much. – Sunit das Nov 22 '20 at 21:38
  • By the theorem in the linked dupe, the map $,x\to ax,$ is injective ($1$ to $1$) $\iff$ its surjective (onto) $\iff \gcd(a,n) = 1\ \ $ – Bill Dubuque Nov 23 '20 at 02:10

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Hint

Suppose there exists $i,j \in \{1,2,\cdots,n-1\}$ such that
$(a \times i) \equiv (a \times j) \pmod{n}.$

What does this say about whether $n$ divides $a(i - j)$?

Of what relevance is the fact that $a$ relatively prime to $n$?

With $i \neq j$, is it possible for $n$ to divide $(i - j).$

user2661923
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