In Abelian group of finite order n, a1*...*an = multiplying set of the elements of order 2.

Question

Let G be a finite Abelian group such that G = { a1,,,an }

And I need to prove that ar*...*an = multiplying set of the elements of order 2.

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I am not sure about the way of approaching it:

1. trivial case ( each of the ai ) where 1 <= i <= n is of O(ai) = 2, therefore sinch G is abelian, by commutative property there are equal.

2. otherwise, there exists at least 1 ai such that O(ai) != 2, which means that the inverse of ai does not belong to the multiplying set either.

3. what if there are no elements of order 2 at all? ( is that even possible? )

Much appreciation

Answer 1

$a_i$ is of order $2$ iif $a_i=a_i^{-1}$. So in your product, you can make 3 packs :

1. the pack of elements of order $1$ : there is only one such element, the unity;
2. the pack of elements of order greater or equal than $3$ : in this pack, every element is with its inverse, so the product of all those elements is unity;
3. finally, the pack of elements of order $2$, whose product is the only one remaining.

One final remark : if there are no order $2$ element (which is, from what precedes, only possible if $n$ is odd), then by definition the product of an empty set of elements is unity, so no harm done.