Proof: Group G of Order 2n Has Elements of Order 2

[astronut24]

if G is a group of order 2n then show that it has an element of order 2 ( and odd number of them)
i've been thinking about this...and i think I've gotten somewhere...
e belongs to G and o(e) = 1
now if there's no element of order 2 in G...we're looking at elements which are not their own inverses...such elements come in pairs, so i guess there lies the contradiction...
i don't have a clue as to how to show that there are odd number of elements of order two.

 

[matt grime]

just carry on in the same vein.

G as a set is the union of the pairs {x,x^{-1}} where these are distinct, with the set of elements of order 2 and the set containing the identity. equating orders

|G| =2n = 2K + L + 1

where K is the number if pairs, and L is the number of self inverse non-identity elements.

 

[M98Ranger]

Your reasoning is correct. Since G has order 2n, it must have at least one element of order 2 (since e has order 1). Now, if there are no other elements of order 2, then all the remaining elements must have order 1, which means they are their own inverses. This creates a pairing of elements where each element is its own inverse, which is not possible in a group (except for the identity element). Therefore, there must be at least one more element of order 2, giving us an odd number of elements of order 2 (1 element of order 2 + an odd number of elements paired with themselves). This proves that G has at least one element of order 2 and an odd number of elements of order 2.