# Thread: Proving an abelian group properties

1. ## Proving an abelian group properties

Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

just a bit stuck on this question, i know lagrange's theorem is involved and have applied it but im just not sure how to finish off the proof, can anyone show me?

thx

2. Originally Posted by TrueTears
Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

just a bit stuck on this question, i know lagrange's theorem is involved and have applied it but im just not sure how to finish off the proof, can anyone show me?

thx
This only holds if n is coprime to 2.

Personally, I would note that there is an element of order 2 in there (why?) and one can quotient out the subgroup generated by it (why?).

An element has order 2 is and only if it is in the kernel of this homomorphism and non-trivial (why?). This proves the result.

3. a somewhat elementary way, is to observe that we have 2n-1 non-identity elements. by pairing each such element with its inverse,

we wind up with an odd number of elements such that x^-1 = x. so we must have at least one element of order 2 (because 0 is not an odd number).

can we have two? well if we did, say, x and y, since G is abelian {e,x,y,xy} is a subgroup of order 4. but 4 does not divide 2n.