Show there exist an element of order 2 in this group

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SUMMARY

If G is a finite group with an even order, then there exists an element a in G such that the order of a is 2. This means that a squared (a^2) equals the identity element of the group. The discussion references Cauchy's Theorem, which supports the conclusion that the existence of such an element implies that a is its own inverse. Furthermore, if no other elements possess this property, the remaining elements of the group can be paired with their inverses.

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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.
 

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