Answer to Question #138068 in Abstract Algebra for Ram

Question #138068
If G is a group of even order, prove that it has an element 'a' which is not equal to 'e' satisfying a^2=e. e is identity element.
1
Expert's answer
2020-10-13T19:09:26-0400

Let G=2n|G|=2n. If a2=ea^2=e, then a2a1=ea1a^2a^{-1}=ea^{-1}, and consequently a=a1a=a^{-1}. By analogy the equality a=a1a=a^{-1} implies aa=aa1aa=aa^{-1}, and thus a2=ea^2=e. So a2=ea^2=e if and only if a1=aa^{-1}=a. Suppose that a2ea^2\neq e for any aea\neq e. Then {a,a1}=2|\{a,a^{-1}\}|=2 for any aea\neq e, and consequently the set G{e}G\setminus\{e\} is partitioned into pairs of inverses elements. This imlies that the set G{e}G\setminus\{e\} has even order. Therefore, the group GG has odd order. This contradiction proves the statement,


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment