I just wanna know if the following proof is okay for my assignment problem.
Question: Prove that an abelian group with 2 elements of order 2, must have a subgroup of order 4.
Proof: Let a,b belong to G such that a2 = b2 = e.
Case:1 G is finite
Now, I propose that there should atleast one more element,x in G.
This is true according to a corollary of Lagarage's theorem which states that,
"In a finite group, the order of the an element of the group divides the order of the group"
So, o(G) = 2n (n>1)
Case 1a Let's first prove there is one more element in G.
We know a2 = b2 = e. Consider the order of ab.
(ab)2 = a * b * a *b = a* a * b *b = a2 * b2 = e.
So, the other element in the group, say x is the product of a and b.
So, G = {e,a,b,x}, When x is denoted as c, this becomes the Klein's 4 group, eh?
Now, every group is a subgroup of itself, the result is proved.
Case 1b Let's suppose there are more than 4 elements in G. Let o(G) = 2n (n>2)
Still, this set H = {e, a, b, ab} satisfies all the necessary conditions for becoming a subgroup of G which proves that G still has a subgroup of order 4.
In addition, i believe this also proposes that the order of G is 4n (n>=1), as any other order would be a contradicting the Lagrange's theorem.
I'm a little confused about the proof for Case 2, where G is of infinite order. We cannot use the corollary due to Lagrange's theorem. Can someone help me with that?