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I am reading the Wikipedia article entitiled Sylow theorems. This short segment of the article reads:

Part of Wilson's theorem states that $(p-1)!$ is congruent to $-1$ (mod $p$) for every prime $p$. One may easily prove this theorem by Sylow's third theorem. Indeed, observe that the number $n_p$ of Sylow's $p$-subgroups in the symmetric group $S_p$ is $(p-2)!$. On the other hand, $n_p ≡ 1$ mod p. Hence, $(p-2)! ≡ 1$ mod $p$. So, $(p-1)! ≡ -1$ mod $p$.

I do not understand why the number of Sylow p-subgroups of $S_p$ is $(p-2)! $

I am taking a course in Group Theory and I have studied everything in Dummit and Foote up to Sylow's theorem. I have also had an introductory course in number theory and am familiar with basic combinatorics.

Rushabh Mehta
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Geoffrey Critzer
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1 Answers1

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The elements of order $p$ consist of a $p$-cycle of the form $(1,a_2,a_3,\ldots,a_p)$, where $a_2,a_3,\ldots,a_p$ is a permutation of $2,3,\ldots,p$. So there are exactly $(p-1)!$ elements of order $p$.

Now each subgroup of order $p$ contains $p-1$ elements of order $p$ (i.e. its non-identity elements), and the intersection of any two such subgroups is trivial, so the total number of subgroups of order $p$ is $(p-1)!/(p-1) = (p-2)!$.

Derek Holt
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  • This is one line more than the solution of "see page number $2$" (which is good, of course). – Dietrich Burde Nov 14 '15 at 20:06
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    And it saves one click! – Derek Holt Nov 14 '15 at 20:14
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    OK, Thanks. I see now that generally the number of cyclic subgroups of order m in S_n is binomial(n,m)*(m-1)!/phi(m) where phi(m) is the number of positive integers less than m that are relatively prime to m. – Geoffrey Critzer Nov 14 '15 at 21:35
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    Unfortunately it is not as simple as that. For example, when $n=m=6$, in addition to elements like $(1,2,3,4,5,6)$, you also have those like $(1,2)(3,4,5)$. – Derek Holt Nov 14 '15 at 22:29
  • Yes, Thank you Derek Holt. I agree. What I think we can say is this: The number of subgroups in S_n of order p (where p is a prime dividing n) is binomial(n,p)*(p-2)!. – Geoffrey Critzer Nov 15 '15 at 17:27
  • @Geoffrey Critzer I'm not sure that can be said either. Suppose $p=2$. Then the number of subgroups of order $2$ is the same as the number of involutions on $[n]$ which is equal to $\displaystyle \sum_{i=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2i}(2i-1)!!$ which is nowhere close to $\binom{n}{2}$. – Aryaman Jal Apr 14 '20 at 16:05
  • Why the intersection of two subgroup must be trivial? – eraldcoil Jul 29 '20 at 05:20
  • @eraldcoil Each subgroup is cyclic. Suppose $P_1\cap P_2\ni a\neq e$. Then $\langle a\rangle =P_1, P_2$. So the subgroups would be equal. – Adam Martens Sep 04 '20 at 17:47