To show: thereexist some x in G such that order of x is 2.

Let \(\displaystyle{S}={\left\lbrace{x}\in{G}{\mid}{\left|{x}\right|}{>}{2}\right\rbrace}\)

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

\(\displaystyle\frac{{G}}{{S}}={\left\lbrace{x}\in{G}{\left|{x}={e}{\quad\text{or}\quad}\right|}{x}{\mid}={2}\right\rbrace}\)

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.

Let \(\displaystyle{S}={\left\lbrace{x}\in{G}{\mid}{\left|{x}\right|}{>}{2}\right\rbrace}\)

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

\(\displaystyle\frac{{G}}{{S}}={\left\lbrace{x}\in{G}{\left|{x}={e}{\quad\text{or}\quad}\right|}{x}{\mid}={2}\right\rbrace}\)

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.