Proof of Euler's Theorem without abstract algebra?

Solution 1:

Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.

Consider a number $c < n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.

First observe that for any $a_i$, $c a_{i} \equiv a_{j} \pmod{n}$ for some $j$. (True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$. This follows immediately from the definition).

If $c a_{i} \equiv c a_{j} \pmod{n}$ then $a_i = a_j$. (True as cancellation can be done since $c$ is relatively prime to $n$).

Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.

Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get

$$c^{\varphi(n)} \equiv 1 \pmod{n}$$ whenever $(c,n) = 1$.