It's about time y'all!!! I figured it out last night while trying to fall asleep.
In case you missed it ('cause I'm not doing all those superscripts and subcripts again!), I blogged about an abstract algebra problem that previously had me stumped. The answer turned out to be something that seems moderately easy to me in hindsight. So yeah, I was both excited AND aggravated at the same time when I finally figured out the solution. That happens with math, though. :-)
I just had to approach the problem from a different angle and think slightly outside the box. I was so focused on proving closure for composition and inverses that I didn't think to check other group properties. When proving that subgroups are groups, you're only required to show closure under operation and inverses. Last night I decided to consider the identity element (All groups have an identity element). With permutations, the identity element is a permutation that carries x to x. All elements are fixed. In effect, e(x)= x for all x in a set A. In problems #1-3, I verified that e existed. Let me break it down problem by problem:
In #1, there's a fixed element a in all the permutations in G such that f(a)= a, and A is not defined as finite or infinite. The identity permutation e can exist in G because a is fixed for all f contained in G.
In #2, there are a finite number of moveable elements and an infinite number of fixed elements (since A is infinite). Finite can mean zero. Zero moveable elements combined with an infinite number of fixed elements corresponds to the identity e. At this point, I started to feel like I was going somewhere with my train of thought.
In #3, B is a subset of a finite set A; and for all x in B, f(x) is also in B. Since A is finite, then B is automatically finite. The elements of B can be fixed and/or moveable when permutated. If fixed, then it's problem #1 all over again. If there are moveable elements in A, there are a finite number of them. That finite number must be zero for e to exist. Since that possiblity exists, then G is still a group.
And now, the counterexample:
In #4, A is infinite. Therefore B could be either finite or infinite. If finite, then we can prove that G is a group using the previous problems. B has always been a finite set in the previous problems, fixed or not. However, if B is infinite and contains only moveable elements, then G cannot contain the identity. With no possiblity of there being all fixed elements in B, e does not exist. That's it. Thank goodness that's over with!