I know it seems obvious, but you should check that . What you have shown is that has order 1 or 2.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.

Is there anything wrong with your particular argument when is any other (possibly infinite) abelian group? Just use the fact that to get rid of your restriction on the size of .Case 1a Let's first prove there is one more element in G.