From A Classical Introduction to Modern Number Theory by Ireland and Rosen, page 33:
Corollary 1 (Euler's Theorem). If $(a,m) = 1$, then $a^{\phi(m)} \equiv 1\,(m)$.
Proof. The units in $\mathbb{Z}/m\mathbb{Z}$ form a group of order $\phi(m)$. If $(a,m) = 1$, $\bar{a}$ is a unit. Thus $\bar{a}^{\phi(m)} = \bar{1}$ or $a^{\phi(m)} \equiv 1\,(m)$.
If I'm interpreting this correctly, this proof implicitly uses the fact(?) that if $G$ is a group and $x\in G$, then $x^{|G|}=1$, where $1$ is the identity element of $G$.
If this is indeed true, can someone explain why? (I have had no prior exposure to group theory.)
Best Answer
Yes, $x^{|G|} = 1$ is correct for any finite group $G$ and any $x \in G$.
(Lagrange's Theorem) If $H$ is a subgroup of $G$, then $|H|$ is a divisor of $|G|$.
If $x \in G$, then there is a smallest positive integer $k$ such that $x^k = 1$. This number $k$ is called the order of $x$.
If $k$ is the order of $x$, then $\{1, x, x^2, \ldots, x^{k-1}\}$ is a subgroup of $G$ having $k$ elements. By step 1, $k$ is a divisor of $|G|$, so $kl = |G|$ for some integer $l$.
$x^{|G|} = x^{kl} = (x^k)^l = 1^l = 1$.