# Statement(s) > [!NOTE] Lemma > Let $G$ be a [[Finite Group]]. Let $g \in G$. Then the [[Order of Group Element|order]] of $g$ divides the [[Cardinality|cardinality]] of $G$. # Proof(s) **Proof 1:** By [[Element of a finite group is of finite order]], $g$ has finite order $n$. Consider the subgroup generated by $g$ hence $\langle g \rangle = \{ 1,g,g^{2},\dots, g^{n-1} \}.$This contains $n$ elements. If $G = \langle g \rangle$ then we are done because $|G| = n$. If not we can find an element $x \in G$ such that $x \not \in \langle g \rangle$. For such an element $x$ consider the set $X = \{ x, xg,xg^{2},\dots, xg^{n-1} \}$We will show that $X$ contains n elements by showing that all the listed elements are distinct. We will also show that $X \cap \langle g \rangle = \emptyset$. Suppose $xg^{i} = xg^{j}$ for integers $i,j$ with $0\leq i<j \leq n-1$. Pre-multiplying both sides by $x^{-1}$ gives $g^{i}=g^{j}$. This is a contradiction because $g^{i}$ and $g^j$ are distinct. Now suppose $X \cap \langle g \rangle \neq \emptyset$. Then there are integers $i,j$ between $0$ and $n-1$ inclusive such that $g^{i} = xg^{j}$ This gives $g^{i-j} = x$ which gain implies that $x \in \langle g \rangle$, a contradiction. Suppose $G = \langle g \rangle \cup X$. Then $G$ has $2n$ elements and the order of $g, n,$ divides the order of $G$. If not we can choose an element $y \in G$ such that $y \not \in \langle g \rangle \cup X$. In a similar way to before with $X$ and $x$ let $Y= \{ y, yg, yg ^{2},\dots, yg^{n-1} \}$Suppose $G = \langle g \rangle \cup X \cup Y$. Then $G$ has $3n$ elements. If not we can continue this process. **Proof 2:** This is a consequence of [[Lagrange's theorem (on Finite Groups)|Lagrange's theorem]] since [[Order of Group Element Equals Order of Subgroup Generated by Element]]. $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography