Element to the Power of Group Order

Theorem

Let $G$ be a group whose identity is $e$ and whose order is $n$.

Then:

$\forall g \in G: g^n = e$

Proof

Let $G$ be a group such that $\left|{G}\right| = n$.

Let $g \in G$ and let $\left|{g}\right| = k$.

From Order of Element Divides Order of Finite Group, $k \backslash n$, and so $\exists m \in \N^*: k m = n$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g^n\)	\(=\)	\(\displaystyle \left({g^k}\right)^m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 25$: Theorem $25.7$