Prove that a group of even order must have an element of order 2.

To show: thereexist some x in G such that order of x is 2.
Let ${\displaystyle S=\{x\in G\mid \left|x\right|>2\}}$
Let ${\displaystyle l\in S}$
Since order of any element in group is equal to order of its inverse, thus ${\displaystyle \left|l^{-1}\right|=\left|l\right|>2}$
Hence, ${\displaystyle l^{l\in S}}$
Since, $|l|>2$ it implies that ${\displaystyle l\ne l^{-1}}$. Hence every lement in S can be paired off with its inverse which is also on S. Thus, S otains even number of elements
${\displaystyle \frac{G}{S}=\{x\in G|x=e\text{or}|x\mid =2\}}$
Since , identity element is unique in a group, therefore G must have an element of order 2 in order to ${\displaystyle \frac{G}{S}}$ with even number of elements.
Therefore, a group of even order must have an element of order 2.