Problem: G = {a1,a2,a3,..,an} is a finite abelian group where order of each element (except of identity of-course) is 2.

Prove

1. Order of G is of the form $\displaystyle n = 2^k$ where k is a positive integer

2. a1.a2.a3....an = e (This is true for all values of n>2.)

My attempt for 1

If the group G, has apropersub-group of order $\displaystyle 2^m$ then it has a sub-group (may not be proper) of order $\displaystyle 2^{m+1}$. This is how I did it. Let H be apropersub-group of order $\displaystyle 2^m < n$. Thus, $\displaystyle \exists$ b $\displaystyle \in G$ but b not $\displaystyle \in H$. Consider $\displaystyle H_1= H \bigcup Hb$ where Hb is a coset of H. $\displaystyle H_1$ is a sub-group of order $\displaystyle 2^{m+1}$. Hence the result as there is a trivial sub-group of order 1 in G.

Any comments on this approach - is there a better way?

Any hints on 2?