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The theorem quoted in the title, was actually stated differently in the problem I was reading. The original statement was as follows:

Let $m$ be a positive integer and let $S$ denote the set of positive integers less than $m$ that are relatively prime to $m$. Prove that for each $x$ in $S$, there exists a unique $y$ in $S$ such that $xy$ is congruent to $1$ modulo $m$.

The proof that I have encountered, addresses the statement of the theorem given in the title:

Consider the sequence of $m$ numbers $0, x, 2x, \ldots, (m−1)x$. We claim that these are all distinct modulo $m$. Since there are only $m$ distinct values modulo $m$, it must then be the case that $ax = 1 \mod m$ for exactly one $a$ (modulo m). This $a$ is the unique multiplicative inverse. To verify the above claim, suppose that $ax = bx \mod m$ for two distinct values $a,b$ in the range $0 \le a,b \le m−1$. Then we would have $(a−b)x = 0 \mod m$, or equivalently, $(a−b)x = km$ for some integer $k$ (possibly zero or negative). But since $x$ and $m$ are relatively prime, it follows that $a−b$ must be an integer multiple of $m$. This is not possible since $a,b$ are distinct non-negative integers less than $m$.

As far as I can understand, this only proves that $x$ always has a unique multiplicative inverse, but not that this inverse belongs to the set $S$ (as defined by the original statement of the theorem).

I understand that this proof is correct, and I can see why it would work when $m$ is prime (as the set $S$ would then contain all positive integers less than $m$), however when $m$ is any positive integer the set $S$ would not necessarily contain $m-1$ elements.

Therefore, it seems as if the proof does not exclude the possibility that the multiplicative inverse is not itself relatively prime to $m$.

JB-Franco
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F. Munnelly
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1 Answers1

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It sounds like you accept that for any $x\in S$ s that there exist a $y\in \{1,\cdots, m-1\}$ such that $xy\equiv 1 \pmod m$ but you are not convinced that $y\in S$

what is then $\gcd(y,m)$?

We know that $xy - pm= 1$ (for some integer $p$) therefore $\gcd(y,m) = 1$

user317176
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  • Is the fact that xy - pm = 1 ⇒ gcd(y, m) = 1 based on a theorem itself? I think it makes sense since if y and m shared a factor k greater than 1, then that could be factored out so that k(xq - pr) = 1, which would imply that k = 1 (along with xq-pr =1) anyway. – F. Munnelly Jul 19 '19 at 18:50
  • It falls out from the application of the Euclidean algorithm. – user317176 Jul 19 '19 at 19:12
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    @F.Munnelly It's the trivial direction of the Bezout identity (when stated in $!\iff!$ form). But it is an immediate consequence of divisibility laws: $,d\mid y,m,\Rightarrow, d\mid xy,pm,\Rightarrow, d\mid xy+pm = 1.,$ Equivalently, sets of (common) multiples $M$ are closed under both addition, and multiplication by any integer, therefore $,y,m\in M,\Rightarrow, xy+pm\in M,$ for any $x,p\in\Bbb Z\ $ (here $M = d\Bbb Z = $ all multiples of $d).,$ Such sets of (common) multiples are prototypical examples of ideals. – Bill Dubuque Jul 19 '19 at 19:25
  • @DougM One doesn't need the Euclidean algorithm or Bezout to deduce that trivial direction - see above. – Bill Dubuque Jul 19 '19 at 19:26