Cauchy's theorem in group theory: stronger statement? 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 $p\mid |X|$, 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.
 A: Note that for any group $G$ and any positive integer $n,$ the number of elements of order $n$ in $G$ is either infinite, or a multiple of $\phi(n),$ where $\phi$ is Euler's $\phi$-function. For we may define an equivalence relation $\sim$ on the set of elements of order $n$ in $G$ via $x \sim y$ if and only if $\langle x \rangle = \langle y \rangle$. Each equivalence class contains $\phi(n)$ elements (assuming the set is non-empty- if it is empty, there is nothing to prove).
A: Frobenius (1895) proved a stronger version of Cauchy's theorem: the number of elements of order p is either 0 or $(p-1)(kp+1)$ for some positive integer k.  For p = 2 this is sharp: the dihedral group of order $4k$ has exactly $(2-1)(2k+1)$ elements of order 2.  For p = 3 however, it is no longer sharp, and you can see one of the medium difficulty cases in this question where we show that k = 15 is not possible.
Frobenius's theorem has a very easy to read proof in Isaacs–Robinson (1992), where we only need up to Theorem 4.
A: All that you have proven is that there are $p$ elements whose order divides $p$. But in general, your claim that there are $p$ elements of order $p$ is false. For example, consider the cyclic group on $p$ elements, which has $1$ element of order $1$ and $p-1$ elements of order $p$.
