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I was watching Michael Penn's video on proof of Euler's Theorem. Here is the link: https://youtu.be/ijT3pmmal00?si=xCe1Rlx_nKHQDoKG

I understand up to the point where he proved that

  1. $GCD(ax_i, n) = 1$
  2. No 2 elements in the set $aS$ are congruent to each other mod n.

But I do not understand why he concluded that $S \equiv aS \pmod{n}$ and the reason why elements of $aS$ $\pmod{n}$ has the same elements of $S$ but just in a different order. In other words, why is an element of $S$ not different from $aS \pmod{n}$ and what makes it?

I figured it has probably something to do with $GCD(a, n)= 1$ but I'm not sure.

Please note that I am a beginner at number theory so please elaborate clearly.

Bill Dubuque
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Hayst
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  • I would say this is a combination of three facts: (1) $aS$ is a subset of $S$; (2) $aS$ and $S$ have the same number of elements; (3) all the elements of $aS$ are distinct modulo $n$. Do you feel comfortable with these facts? Do you agree that together they imply $aS=S$? – Greg Martin Oct 28 '23 at 08:11
  • @GregMartin $aS$ isn't a subset of $S$ for most $a$. – TC159 Oct 28 '23 at 08:30
  • @Hayst In the video we see that claim 1 is just about this, that $aS\subset S$ for those $a$ for which $\gcd(a,n)=1$ . – kabenyuk Oct 28 '23 at 10:05
  • I just dont get why after we reduce mod n to aS, S = aS (mod n). I get the the two points from above. – Hayst Oct 28 '23 at 11:15
  • Okay. I get it now. When we reduce mod n to aS, we know 0 <= aS < n. We also know the elements of aS are all coprime to n, and also incongruent to each other modulo n. This is quite similiar to our definition of S. We define S = {1 <= x < n, GCD(x, n) = 1}. But notice that elements of aS mod n cannot be 0. Ultimately, the conditions of aS becomes 1 <= x < n. Which is why S ≡ aS (mod n). – Hayst Oct 28 '23 at 11:26
  • The same proof of the Theorem (in dupe) shows the scaling by $,a,$ map is a bijection (permutation) of $,S,,$ since elements of $S$ are closed under product. – Bill Dubuque Oct 28 '23 at 19:04

1 Answers1

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Let me phrase this one precisely: we have the set $ S = \{x_i\} $ of all residues modulo $ n $ which satisfy $ GCD(x_i, n) = 1 $, meaning each residue is coprime with $ n $. Then we have another set $ aS $, which has the same number of elements — this number is equal to $ \varphi(n) $ — and satisfies the following conditions:

  1. Each element is coprime with $ n $;
  2. Every two elements of the set have different residues modulo $ n $.

The claim Michael is making goes like this: it follows from the facts stated above that elements of $ aS $ should have exactly the same residues modulo $ n $ as $ S $ does, i.e. those residues which are coprime with $ n $.

Why is that?

I claim that we need to prove the following: $ 0 \leq x < n $ is coprime with $ n $ if and only if $ nk + x $ is coprime with $ n $, where $ k \in \mathbb{N} $. That would mean that when we are talking about 'comprimeness' with $ n $, we only need to consider the residue modulo $ n $ and check if it's coprime with $ n $ or not.

I will do one part, the other one is pretty similar.

So, let's prove that if $ x $ is coprime with $ n $, then $ nk + x $ (where $ k \in \mathbb{N} $ and $ 0 \leq x < n $) is coprime with $ n $. Suppose that $ nk + x $ is not coprime with $ n $, meaning there is a number $ d > 1 $ which divides both $ nk + x $ and $ n $. But that would mean that $ d $ also divides $ (nk + x) - nk = x $, since if $ d $ divides $ n $, it surely divides $ nk $, and if some $ z $ divides $ k $ and $ m $, then it divides $ k - m $. It follows that $ d $ divides both $ n $ and $ x $, meaning $ x $ and $ n $ are not coprime, which is a contradiction.

Assuming we proved the opposite implication (if $ nk + x $ is coprime with $ n $, then $ x $ is coprime with $ n $), we can now prove Michael's claim.

We have two sets of numbers of equal sizes $ \varphi(n) $, both of which contain numbers with different residues modulo $ n $ and are also coprime with $ n $. Using the claim we proved, we can say that if we consider those sets modulo $ n $ (i.e. every number is replaced with its residue modulo $ n $), the properties above would also hold (specifically, the 'coprimeness'). Now those sets modulo $ n $ are both of the size $ \varphi(n) $ and contain different residues modulo $ n $, which are coprime with $ n $. But there are only $ \varphi(n) $ residues modulo $ n $, which are coprime with $ n $. This means that these are precisely the residues contained in our sets modulo $ n $, so these sets are equal modulo $ n $.

Dmitry Ivanov
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