# Eulers Theorem

2021 Apr 15
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This is euler's theorem (for coprime n and k)

$k^{\phi(n)} \equiv 1 \pmod n$

## Key fact

The integers coprime to n, mod n forms a group.
since multiplication by $k$ is invertible (since k is coprime to n, and k is in the group). Multiplying each element by $k$ **just reorders the elements**.

Example: multiply each coprime number by $3$ mod $8$ (Remember mod is the same as the remainder from divison)

$\begin{aligned}
(1 \cdot 3) &= 3 &\to 3 \\
(3 \cdot 3) &= 9 &\to 1 \\
(5 \cdot 3) &= 15 &\to 7 \\
(7 \cdot 3) &= 21 &\to 5
\end{aligned}$

And the product is the same if each element is multiplied by $3$, since we take out as many $8$'s as we can from each term.

$1 \cdot 3 \cdot 5 \cdot 7 \equiv (3 \cdot 1) (3 \cdot 3) (3 \cdot 5) (3 \cdot 7) \equiv 3 \cdot 1 \cdot 7 \cdot 5 \pmod 8$

## The Proof

If we have $\{k_1, \dots, k_{\phi(n)}\}$ being the integers coprime to n.

We know $k$ is in the set (since n and k are coprime).
if we multiply each $k_i$ by $k$ we have

$\prod^{\phi(n)} k_i \equiv \prod^{\phi(n)} k_i k \pmod n$

Because of invertibility (no overlaps/no duplicates) and closure (stays in the set of coprimes), multiplying each $k_i$ by $k$ just reorders the elements.

Now we factor out k and invert/divide by the product (which will just be another k)

$1 \cdot \bcancel{\prod^{\phi(n)} k_i} \equiv k^{\phi(n)} \bcancel{\prod^{\phi(n)} k_i} \pmod n$

So we end up with

$k^{\phi(n)} \equiv 1 \pmod n$

## Proof it is a group

There are 4 group axioms we must satisfy

- Closure: if $k_1$ and $k_2$ share no factors with $n$, $k_1 k_2$ share no factors with $n$.
- An identity element: just $1$ in our case
- Associativity: multiplication is associative
- Invertibility: see below

Let $k$ be an element of the group,
we know that $\gcd(k, n) = 1$ (coprime)

Euclids algorithm implies there exist $s$ and $t$ such that

$sk + tn = 1$

(Also known as Bézout's identity)

Now mod out by $n$

$sk + tn = 1 \implies sk \equiv 1 \pmod n$

Therefor $s = k^{-1}$, and an inverse exists!