TITLE: Prove a group of order 12 must have an element of order 2
QUESTION [8 upvotes]: Question: Prove that a group of order 12 must have an element of order 2.

I believe I've made great stride in my attempt.
By corollary to Lagrange's theorem, the order of any element $g$ in a group $G$ divides the order of a group $G$.
So, $ \left | g \right | \mid \left | G \right |$.
Hence, the possible orders of $g$ is $\left | g \right |=\left \{ 1,2,3,4,6,12 \right \}$
Suppose $\left | g \right |=12.$
Then, $g^{12}=\left ( g^{6} \right )^{2}=e.$
So, $\left | g^{6} \right |=2$
Using the above same idea and applying it to $\left | g \right |=\left \{ 6,4,2 \right \}$ and $\left | g \right |=1,$
we see that these elements g have order 2.
However, for $\left | g^{3} \right |$, the group $G$ does not require an element of order 2.
How can I take this attempt further?
Thanks in advance. Useful hints would be helpful.

REPLY [15 votes]: Hint: Here is a simple proof idea that every group of even order must have an element of order $2$.
Pair every element in $G \backslash \{ e \}$ with its inverse. If all pairs consist of two different elements then $G \backslash \{ e \}$ would have an even number of elements. 
What does it mean that $a=a^{-1}$?

 $a=a^{-1} \Leftrightarrow a^2=e$. And since $a \neq e$ we get that $ord(a)=2$.