Proof: Order of an Element Divides the Order of a Group

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Problem

If $G$ is a finite group, prove that the order of any element of $G$ divides the order of $G$.

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Definitions Used

1. Finite Group: A group $G$ with a finite number of elements, called the order of the group, denoted $|G|$.
2. Order of an Element, $o(a)$: For an element $a \in G$, its order is the smallest positive integer $m$ such that $a^m = e$.
3. Cyclic Subgroup, $\langle a \rangle$: The subgroup generated by an element $a$, consisting of all integer powers of $a$. The order of this subgroup, $|\langle a \rangle|$, is equal to the order of the element, $o(a)$.
4. Lagrange’s Theorem: The order of any subgroup of a finite group divides the order of the group.

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Resolution

Let $G$ be a finite group, and let its order be $n$. So, $|G|=n$.

Let $a$ be an arbitrary element in $G$, and let its order be $m$. So, $o(a)=m$. Our goal is to prove that $m$ divides $n$.

Consider the cyclic subgroup $H$ generated by the element $a$. This subgroup consists of all the distinct powers of $a$:

$$ H = \langle a \rangle = \{e, a, a^2, \dots, a^{m-1}\} $$

The number of distinct elements in this subgroup is $m$. Therefore, the order of the subgroup $H$ is $m$.

$$ |H| = m $$

According to Lagrange’s Theorem, the order of any subgroup of a finite group must divide the order of the group itself. Since $H$ is a subgroup of $G$, we must have:

$$ |H| \text{ divides } |G| $$

Substituting the orders we defined, we get:

$$ m \text{ divides } n $$

Conclusion

We have shown that the order of the element $a$ (which is $m$) divides the order of the group $G$ (which is $n$). Since $a$ was an arbitrary element, this holds true for every element in the group.