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

Prove that a group of even order must have an element of order 2.
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joshyoung05M

To show: thereexist some x in G such that order of x is 2.
Let $S=\left\{x\in G\mid |x|>2\right\}$
Let $l\in S$
Since order of any element in group is equal to order of its inverse, thus $|{l}^{-1}|=|l|>2$
Hence, ${l}^{l\in S}$
Since, $|l|>2$ it implies that $l\ne {l}^{-1}$. Hence every lement in S can be paired off with its inverse which is also on S. Thus, S otains even number of elements
$\frac{G}{S}=\left\{x\in G|x=e\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}|x\mid =2\right\}$
Since , identity element is unique in a group, therefore G must have an element of order 2 in order to $\frac{G}{S}$ with even number of elements.
Therefore, a group of even order must have an element of order 2.