Help understanding two theorems on primes and residue systems

Question

My textbook lists two theorems and I'm not sure how I'm supposed to interpret them. I don't need a proof; I'm only trying to figure out what information I'm being told by each theorem.

Let $p$ be a prime and let $a$ be an integer not divisible by $p$; that is, $gcd(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$

So for this first theorem, I believe the canonical complete residue system $p$ would be $\{0,1,2,...,p-1\}$. So the set $\{a,2a,3a,...,pa\}$ would be the integers that satifsy the congruences, $a\equiv0 \space (mod \space p), 2a\equiv1 \space (mod \space p), 3a\equiv2 \space (mod \space p),...,pa\equiv (p-1) \space (mod \space p)$.

Is my interpretation correct?

Let $p$ be a prime and let $a$ be an integer not divisible by $p$. Then,

$a\cdot2a\cdot3a\cdot...\cdot(p-1)a\equiv1\cdot 2 \cdot 3 \cdot...\cdot(p-1) \space (mod \space n)$

For this second theorem, is this theorem telling me that I can decompose congruence integers into products of primes?

Thanks for any help.

Answer 1

Okay, what does it mean.

A complete residue system $\mod p$. What's that?

If you compare any integer $n \pmod p$ that are only $p$ options. Either $n\equiv 0\pmod p$ or $n\equiv 1 \pmod p$ or $n\equiv 2 \pmod p$ or ...... or $n\equiv p-1\pmod p$.

So the list $\{0,1,2,.....,p-1\}$ is called a complete residue system because it contains a representation, a residue, for every case.

But that there nothing special about choosing just those (the smallest non-negative) residues. We could have chosen any $\{r_1, r_2, ....., r_p\}$ so that every integer is equivalent to one of them (and we can list them in any order).

For example to find a complete residue system $\mod 6$ we can represent all the integers that have remainder of $0$ by $78$. (Because $6|78$). And we can represent all the integers that have a remainder of $2$ with $-22$ (because $-22 = -4*6 + 2$). And we can represent all the integers that have remainder of $5$ with $647$ (because $647 = 642 + 5=6*107 + 5$) and so on.

So $\{78, -22, 647, 3001, -2,10\}$ is a complete residue system $\pmod 6$ because every interger is congruent to exact one of those $\pmod 6$. (test them: $$0\equiv 78\pmod 6; 1\equiv 3001 \pmod 6; 2\equiv -22\pmod 6; 3\equiv -2\pmod 6; 4\equiv 10\pmod 6; 5\equiv 647\pmod 6; 6\equiv 78\pmod 6; 7\equiv 3001\pmod 6$$ and so on. [they just cycle and cycle through].

(Admittedly, it's a fairly irrelevant and awkward way to represent them... but it does represent all $6$ possible congruences.)

So the theorem is saying: If $p$ is a prime and $a$ is a number that .... isn't a multiple of $p$... then the set $\{a, 2a, 3a,.... , pa\}$ is a complete residue system $\mod p$ and that every integer is congruent to exactly one of those number.

For example: $\mod 7$ and the number $3$ we'd have $\{3,6,9,12,15,18,21\}$ is a complete residue system.

And if we test it. $$0\equiv 21 \pmod 7; 1\equiv 15\pmod 7; 2\equiv 9\pmod 7; 3\equiv 3 \pmod 7; 4\equiv 18\pmod 7; 5\equiv 12\pmod 7; 6\equiv 6\pmod 7; 7\equiv 21 \pmod 7; 8\equiv 15\pmod 7$$ and so on....

We could to it with any number (as long as it's not a multiple of $7$). For example...... $93$....

$\{93, 186, 279, 372, 465, 558,651\}$ should work. Does it? I'm sure it does because I have complete faith in the theorem but looking at that I'd have no idea if it weren't for the theorem but:

$651 = 7*93$ that was a gimme; $372=53*7 + 1$ and $93=7*13+2$ and $465=7*67+3$ and $186=7*26+4$ and $558 = 7*79+5$ and $279=7*39 +6$ and so on.

It does work.

Answer 2

No, you have misunderstood both. Let's consider a simple special case: $\,a=2,\, p=5.\,$ i.e. we consider the doubling map $\,n\to 2n\bmod 5\,$ whose action is as follows.

$\qquad\qquad\bmod 5\!:\,\ \begin{align}&2\{\color{#c00}1,2,3,\color{#0a0}4\} \\ \equiv\, &\ \ \{\color{#c00}2,4,1,\color{#0a0}3\}\end{align} $

i.e. $\ \ \ \quad\qquad\qquad\qquad\begin{align}2\cdot \color{#c00}1&\equiv\color{#c00} 2\\ 2\cdot 2&\equiv 4\\ 2\cdot 3&\equiv 1\\ 2\cdot \color{#0a0}4&\equiv\color{#0a0}3 \end{align}\,$

Multiplying them all yields product of all on LHS is congruent to product of all om RHS, i.e.

$\quad\qquad\qquad\begin{align}&(2\cdot\color{#c00} 1)(2\cdot 2)(2\cdot 3)(2\cdot\color{#0a0} 4)\\[.1em] \equiv &\ \ \, (\color{#c00}2)\, \cdot\, (4)\,\cdot\, (1)\,\cdot\, (\color{#0a0}3)\\[.1em] \equiv&\ \ \,4!\,\equiv -1\end{align}\,$ by the Congruence Product Rule.

That's what the theorem is telling you (and how it is proved).

Remark $ $ The key idea is: $\,\gcd(a,p)\!=\!1\Rightarrow$ the map $\,x\to ax\,$ is invertible $\!\bmod p\,$ so it is a bijection, so it simply permutes the complete set of nonzero residues $\,1,2,\ldots,p\!-\!1,\,$ which does not alter their product $\,(p\!-\!1)!,\,$ e.g. above the doubling map permuted $\,\color{#c00}1,2,3,\color{#0a0}4\ $ to $\ \color{#c00}2,3,1,\color{#0a0}3$.

The multuiplication is a congruence generalization of the fact that equal finite sets of integers have equal products, by replacing "equal" by "congruent" (congruence can be viewed as a generalized equality - in fact it becomes an equality if we work in the quotient ring $\,\Bbb Z_p\,$ of cosets $\,[k] = k+p\Bbb Z)$

For another common example of this idea see the Wilson reflection formula where the permutation arises from an additive shift by $\,a\,$ (vs. multiplicative scaling by $\,a)$.