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If $n =p$ is a prime number and if $[0]\neq[a]$ is in $J_p$ , then there is an element $[b]$ in $J_p$ such that $[a][b] = [1]$. ($J_p$ being the set of congruence classes mod $p$)

This was a remark in the book Topics of Algebra by I.N. Herstein in the congruence modulo section of the topic Integers. He didn't prove it.

I tried to prove it but I have no clue as to how I'm supposed to approach this, all I could do is simplify this to : Prove, for all $a<p$ there exists a $b<p$ such that $ab=np + 1$ where $p$ is prime and $n (<a,b)$ is any integer.

Could anyone show how this is true or maybe help me out with how to approach it.

Bill Dubuque
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Karan
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3 Answers3

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Let $p$ be prime and let $a, b$ be integers less than $p$. We know from Fermat’s little theorem that $$a^{p-1}=1 (\text{mod} \ p)$$

Thus, we have that $b=a^{p-2} (\text{mod} \ p)$ satisfies this congruence.

Turquoise Tilt
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W M Seath
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  • But how do you know that $a^{p-2}< p$. Also i think you meant $b = a^{p-2}$. – Karan Apr 23 '19 at 07:32
  • You do not need $a^{p-2}\lt p $ to be true, as we are dealing with congruence classes. We have that $a^{p-2}=b+np$ as shown by the modular arithmetic for some integer $n$, so our least positive residue, ie less than p, is simply b. – W M Seath Apr 23 '19 at 08:44
  • Ah I see, thanks. – Karan Apr 23 '19 at 09:39
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You can use the existence of a Bézout identity: since $[a]\neq[0]$, $p$ does not divide $a$, in particular $a$ is prime to $p$ and there exist integers $b$, $q$ such that $ab+pq=1$, for the classes in $J_p$, this implies $[a][b]+[p][q]=[1]$, but since $[p]=[0]$, this yields $[a][b]=[1]$.

gcousin
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If $a=1$ then the result is immediate, so suppose $1<a<p$.

Consider the first $p$ multiples of $a$, which are $a, 2a, 3a, \dots, pa$. Suppose two of these leave the same remainder modulo $p$ - say $ma=na \mod p$ with $m>n$. Then $(m-n)a$ is a multiple of $p$. But since $1<a<p$, this would give a factorization of $p$ which is not the trivial factorization $1 \times p = p$. Since $p$ is prime a non-trivial factorization does not exist. Therefore the first $p$ multiples of $a$ must all leave different remainders modulo $p$.

But there are only $p$ different remainders modulo $p$, so one of the multiples $ka$ with $1<k<p$ must leave remainder $1$ modulo $p$.

gandalf61
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