Proof on the Number of Elements of Order 2
Problem
Show that in a finite group of even order, the number of elements of order 2 is odd.
Definitions Used
- Group of Even Order: A finite group $G$ where the number of elements, $|G|$, is an even integer.
- Element of Order 2: An element $a \in G$ such that $a \neq e$ and $a^2 = e$.
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 grouped into distinct pairs of the form $\{g, g^{-1}\}$. This means that the total number of elements in $A$, denoted $|A|$, must be an even number.
The union of these two disjoint sets is the entire group, so we can write:
$$ |G| = |A| + |B| $$We are given that $|G|$ is even, and we have shown that $|A|$ is even. It follows that $|B|$ must also be even, since $|B| = |G| – |A|$, and the difference between two even numbers is even.
Now, let’s identify the elements contained in set $B$. An element $g$ is in $B$ if and only if $g = g^{-1}$, which is equivalent to the condition $g^2 = e$. This condition is satisfied by exactly two types of elements:
- The identity element, $e$, which has order 1.
- All elements of order 2.
Therefore, the set $B$ consists of the identity element plus all the elements of order 2. We can express the size of $B$ as:
$$ |B| = (\text{the number of elements of order 2}) + 1 $$By rearranging this equation, we can find the number of elements of order 2:
$$ (\text{the number of elements of order 2}) = |B| – 1 $$We already established that $|B|$ is a non-zero even number (it’s non-zero because it must contain $e$). An even number minus 1 is always an odd number.
Conclusion
The number of elements of order 2 is odd in any finite group of even order.