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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.

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You are proving the stronger statement that there are at least $p$ elements of order dividing $p$. And this is equivalent to the usual statement: take an element $g$ of order $p$, and look at the subgroup it generates... – user641 Dec 20 '11 at 6:53
Related to… – lhf Dec 20 '11 at 10:43
up vote 4 down vote accepted

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).

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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.

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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 $p$.

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Yes, i ignored the fact that one of them is identity. Taking that into account, can I say that I am going to have at least p-1 distinct elements of order p (since there can only be one element of order 1)? So I am guranteed the existence of p-1 elements of order p. Probably I am just fussing over nothing here, but since it is not something that jumps out to you ,I find it strange that the theorem would not be worded like that. – Sangeeta Dec 20 '11 at 10:17
@Sangeeta: Yes, the result could be stated in that way, and be then in some sense "stronger." I do not see a great deal of difference, since element of order $p$ and subgroup of order $p$ (with necessarily $p-1$ generators) are so closely connected. – André Nicolas Dec 20 '11 at 19:08

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