Finite Groups, Order of x divides Order of G

Corollary

If $G$ is a finite Group and $x \in G$ , then $| x |$ (see Order (Groups)) divides $| G |$ . In particular, $x^{| G |} = 1$ for all $x \in G$ .

Proof

Using Orders of Cyclic Subgroups, then $| x | = | ⟨ x ⟩ |$ . The first part of the corollary follows from Lagrange's Divisibility Theorem of Order of Subgroups applied to $H = ⟨ x ⟩$ .

The second statement is true since the the previous statement has $x^{| H |} = 1$ , so then:

x^{| G |} = (x^{| H |})^{k} = 1^{k} = 1

for some integer $k$ such that $\frac{| G |}{| H |} = k$ .

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