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# 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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Abstract algebra asked 2021-01-06
Prove that a group of even order must have an element of order 2.

## Answers (1) 2021-01-07
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.

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