In preparation for a test, I have been looking at examples I covered in class and trying to understand them better. I recently did the following example:
If $a^2$ does not equal e for all a, then a does not equal $a^{-1}$, hence this pairing exhausts all non-identity elements. Therefore the number of elements of the group is $2n + 1$, where n is the number of pairings and 1 is the identity element e. But we assumed that G was a group of even order, so this is a contradiction. So G contains an element of order 2.
Does this seem right?
Thanks for reading, greatly appreciate it!
we have $a^2=e$ then $a=a^{-1}$, let $A=\{g\in G; g\ne g^{-1} \}$ then $e\notin A$ ,if $g\in A \Rightarrow g^{-1}\in A$, then we have $|A|$ even number and $|A\cup\{e\}|$ odd number, therefore there exists $a\in G$ and $a\notin A \cup \{e \}$ where $a=a^{-1}$ i.e $a^2=e$