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 a^{2}= b^{2}= 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 a^{2}= b^{2}= e. Consider the order of ab.

(ab)^{2}= a * b * a *b = a* a * b *b = a^{2}* b^{2}= 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?