Proof: Existence of an Element of Order 2
Problem
Show that any finite group of even order must contain an element of order 2.
Definitions Used
- Group of Even Order: A finite group $G$ where the number of elements, $|G|$, is an even integer.
- Order of an Element: The order of an element $a \in G$ is the smallest positive integer $n$ such that $a^n = e$. An element of order 2 is an element $a \neq e$ such that $a^2 = e$.
- Inverse of an Element: For each element $g \in G$, there exists a unique element $g^{-1} \in G$ such that $gg^{-1} = g^{-1}g = e$.
Resolution
Let $G$ be a finite group of even order. We can partition the elements of $G$ into two disjoint subsets:
- Let $A$ be the set of elements that are not their own inverse: $A = \{g \in G \mid g \neq g^{-1}\}$.
- Let $B$ be the set of elements that are their own inverse: $B = \{g \in G \mid g = g^{-1}\}$.
The elements in set $A$ can be paired up, with each element $g$ being paired with its distinct inverse $g^{-1}$. This means that the number of elements in $A$, denoted $|A|$, must be an even number.
Since $A$ and $B$ are disjoint and their union is the entire group $G$, we have:
$$ |G| = |A| + |B| $$We are given that the order of the group, $|G|$, is even. We have also established that $|A|$ is even. Therefore, the number of elements in set $B$ must also be even, because:
$$ |B| = |G| – |A| $$The difference between two even numbers is always an even number.
Now, consider the set $B$. We know that the identity element, $e$, is always its own inverse ($e = e^{-1}$), so $e \in B$. This means that $|B|$ is not zero.
Since $|B|$ is a non-zero even number, it must contain at least two elements. One of these elements is $e$. Therefore, there must exist at least one other element in $B$. Let’s call this element $a$, where $a \neq e$.
By the definition of the set $B$, any element within it is its own inverse. So, for our element $a$, we have:
$$ a = a^{-1} $$If we multiply both sides on the left by $a$, we get:
$$ a \cdot a = a \cdot a^{-1} $$ $$ a^2 = e $$Since $a \neq e$, the order of $a$ is not 1. The equation $a^2=e$ shows that the order of $a$ must be 2.
Conclusion
We have successfully found an element $a$ in $G$ such that $a \neq e$ and $a^2=e$. Therefore, any finite group of even order must contain an element of order 2.