Show there exist an element of order 2 in this group

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In a finite group G with even order, there exists an element a whose order is 2, meaning a^2 equals the identity element. This implies that a is its own inverse. The discussion references Cauchy's Theorem, which supports the existence of such an element. If no other elements had this property, the remaining members of the group could be paired with their inverses, leading to a contradiction. Therefore, the existence of an element of order 2 is confirmed.
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Homework Statement


If G is a finite groups whose order is even, then there exists an element a in G whose order is 2.


Homework Equations





The Attempt at a Solution


does this mean that a^2 is the identity? how can i prove this? Also, would't this make G cyclic?
 
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Yes, that would make a2 = 1
Perhaps you should check out Cauchy's Theorem
 


halvizo1031 said:

Homework Statement


If G is a finite groups whose order is even, then there exists an element a in G whose order is 2/


Homework Equations





The Attempt at a Solution


does this mean that a^2 is the identity? how can i prove this? Also, would't this make G cyclic?
Yes, that means that a^2 is the identity. And that, in turn, means that a is its own inverse.

The group identity has the property that it is its own inverse (but has order 1, not 2). Suppose that there were no other member of the group with that property. Then we could pair the other members of the group, each paired with its inverse.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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