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I will refer to the following simple proof of Cauchy's theorem that appears in chapter 33 of Pinter's A Book of Abstract Algebra. I have copied it below so my question can be properly understood.

This proof is crystal clear, however what I cannot understand is why $p$ has to be a prime number? It seems to me that the proof works for any divisor of $|G|$. Could somebody clarify this? I would appreciate it.


Cauchy's theorem: Let $G$ be a finite group of order $n$ and let $p$ be a prime divisor of $n$, then $G$ has an element of order $p$.

Pinter proves Cauchy's theorem specifically for $p=5$; however, he says, the same argument works for any value of $p$.

Consider all possible 5-tuples $(a, b, c, d, k)$ of elements of $G$ whose product $abcdk =e$.

How many distinct 5-tuples of this kind are there?

Well, if we select $a, b, c$ and $d$ at random, there is a unique $k=d^{-1} c^{-1} b^{-1} a^{-1}$ in $G$ making $abcdk = e$. Thus, there are $n^4$ such 5-tuples.

Call two 5-tuples equivalent if one is merely a cyclic permutation of the other. Thus, $(a, b, c, d, k)$ is equivalent to exactly five distinct 5-tuples, namely $(a, b, c, d, k)$, $(b, c, d, k, a)$, $(c, d, k, a, b)$, $(d, k, a, b, c)$ and $(k, a, b, c, d)$.

The only exception occurs when a 5-tuple is of the form $(a, a, a, a, a)$ with all its components equal; it is equivalent only to itself. Thus, the equivalence class of any 5-tuple of the form $(a, a, a, a, a)$ has a single member, while all the other equivalence classes have five members.

Are there any equivalence classes, other than ${(e, e, e, e, e)}$, with a single member? If not then $5$ divides $(n^4-1)$ (for there are $n^4$ 5-tuples under consideration, less $(e, e, e, e, e)$), hence $n^4\equiv 1\pmod 5$. But we are assuming that 5 divides $n$, hence $n^4\equiv 0\pmod 5$, which is a contradiction.

This contradiction shows that there must be a 5-tuple $(a,a,a,a,a)\neq(e, e, e, e, e)$ such that $aaaaa=a^5=e$. Thus, there is an element $a\in G$ of order 5. ■

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    $\begingroup$ Can you think of a group with order 4, and no element of order 4? $\endgroup$ – Calvin Lin Jul 6 '13 at 0:47
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    $\begingroup$ Calvin Lin's comment hints at an excellent way of working out these questions for yourself. You take a small counterexample of the proposition ($\mathbb Z_2\times\mathbb Z_2$ is the smallest counterexample) and try out the constructions and the claims in the proof against it. $\endgroup$ – Karl Kroningfeld Jul 6 '13 at 2:08
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    $\begingroup$ As others have pointed out, it is the sentence beginning "The only exception" that is only correct when $p$ is prime. Notice that no justification has been included for this assertion, so you should try and prove that it really is correct when $p$ is prime. $\endgroup$ – Derek Holt Jul 6 '13 at 12:54
  • $\begingroup$ Following the reasoning in the proof, if there is a 5-tuple $(a, a, a, a, a) \neq (e, e, e, e, e)$, then $5$ should divide $n^4 - 2$, right? Considering this, can we not say that there are 4 such 5-tuples, so that $n^4 \equiv 0 \pmod 5$? $\endgroup$ – David Cian Mar 10 at 13:07
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"The only exception occurs when a 5-tuple is of the form $(a,a,a,a,a)$ with all its components equal; it is equivalent only to itself."

This breaks if $p = 6$, for example: $(a, a, b, a, a, b)$. But even then:

"there must be a 5-tuple $(a,a,a,a,a)\neq (e,e,e,e,e)$ such that $aaaaa=a^5=e$. Thus, there is an element $a\in G$ of order 5."

This also breaks if $p = 6$: there are plenty of groups $G$ with non-identity elements $a$ such that $a^6 = e$ because $a$ has order 2 or 3 (consider $C_2$ or $C_3$ and a non-identity element).

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Well, one major reason that the proof does not work for composite p is that the 'equivalence class' of a p-tuple being cyclically rotated, may not have either exactly one or exactly p elements in it. For example, if p = 6; look at abcabc.

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