Let "|G|=2n". If "a^2=e", then "a^2a^{-1}=ea^{-1}", and consequently "a=a^{-1}". By analogy the equality "a=a^{-1}" implies "aa=aa^{-1}", and thus "a^2=e". So "a^2=e" if and only if "a^{-1}=a". Suppose that "a^2\\neq e" for any "a\\neq e". Then "|\\{a,a^{-1}\\}|=2" for any "a\\neq e", and consequently the set "G\\setminus\\{e\\}" is partitioned into pairs of inverses elements. This imlies that the set "G\\setminus\\{e\\}" has even order. Therefore, the group "G" has odd order. This contradiction proves the statement,
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