A group of even order contains an odd number of elements of order 2

[mathmari]

Hey!

"Show that a group of even order contains an odd number of elements of order $2$."

We know that the order of an element of a finite group divides the order of the group.

Since, the order of the group is even, there are elements of order $2$.

But how can I show that the number of these elements is odd?? (Wondering)

 

[Euge]

Hey!

"Show that a group of even order contains an odd number of elements of order $2$."

We know that the order of an element of a finite group divides the order of the group.

Since, the order of the group is even, there are elements of order $2$.

But how can I show that the number of these elements is odd?? (Wondering)

Let $G$ be your group, with identity $e$. Let $A$ be the set of all elements of $G$ of order 2. Let $S$ be the complement of $A$ in $G - \{e\}$. Then $S$ consists of all the nonidentity elements $a \in G$ such that $a \neq a^{-1}$. By pairing every nonidentity element with its inverse, we observe that $S$ has an even number of elements. So since $G - \{e\}$ has odd order, it follows that $A$ has an odd number of elements.

 

[Deveno]

Let $G$ be your group, with identity $e$. Let $A$ be the set of all elements of $G$ of order 2. Let $S$ be the complement of $A$ in $G - \{e\}$. Then $S$ consists of all the nonidentity elements $a \in G$ such that $a \neq a^{-1}$. By pairing every nonidentity element with its inverse, we observe that $S$ has an even number of elements. So since $G - \{e\}$ has even order, it follows that $A$ has an odd number of elements.

$G -\{e\}$ has an odd number of elements, since $G$ is of even order.

$|G| = 1 + |A| + |S|$

Since $|G|$ is even, and $|S|$ is even, it follows that:

$|G| - |S| = 1 + |A|$ is likewise even. Thus:

$|A| = |G| - |S| - 1$ is odd.

(It's probably a simple typo, but I thought the OP should be absolutely clear on this).

 

[Euge]

$G -\{e\}$ has an odd number of elements, since $G$ is of even order.

$|G| = 1 + |A| + |S|$

Since $|G|$ is even, and $|S|$ is even, it follows that:

$|G| - |S| = 1 + |A|$ is likewise even. Thus:

$|A| = |G| - |S| - 1$ is odd.

(It's probably a simple typo, but I thought the OP should be absolutely clear on this).

Yes, it was a typo. Thanks, I've made the correction.

 

[blue_raver22]

This can be shown using the theorem that states: In a finite group, the number of elements of order $2$ is always even.

Since the group has even order, there must be at least two elements of order $2$. Let's call them $a$ and $b$.

Now, consider the subgroup generated by $ab$. This subgroup will have order $2$ since $(ab)^2 = a^2b^2 = e$. This means that $ab$ is also an element of order $2$.

But since $a$ and $b$ are already counted as elements of order $2$, the only way for the total number of elements of order $2$ to be even is if $ab$ is equal to one of them.

This means that there is only one element, $ab$, that is not counted yet. Therefore, the total number of elements of order $2$ in a group of even order must be odd.