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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.

Edit: to satisfy 6 character requirement.

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.

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, $np ≡ 1$ 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.

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.

Edit: to satisfy 6 character requirement.

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, np ≡ 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.

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, $np ≡ 1$ 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.