I have seen the following argument being made more than once, for example in the proof of Cauchy's theorem (if G is a finite group and p is a prime number dividing the order of G, then G contains an element of order p. ):
Let $p$ be a prime and X be some set. If $p||X|$, and we know that X contains at least one element, then X has at least p elements.
When we use this argument to prove Cauchy's theorem, are we not proving the much stronger statement that G has at least p elements of order p? Am I missing something here? If not, then why do we not state the theorem in its much stronger form? I have taken care to check that X has distinct elements, so that all p of them could not be the same.
I am using the textbook "A first course in Abstract Algebra" by Fraleigh, and a similar argument is used more than once to prove a couple of different things. A cursory glance at the wikipedia page for Cauchy's theorem shows a similar proof there.