# Explanation of claim in Dummit and Foote

Dummit and Foote, p. 204

They suppose that $G$ is simple with a subgroup of index $k = p$ or $p+1$ (for a prime $p$), and embed $G$ into $S_k$ by the action on the cosets of the subgroup. Then they say

"Since now Sylow $p$-subgroups of $S_k$ are precisely the groups generated by a $p$-cycle, and distinct Sylow $p$-subgroups intersect in the identity"

Am I correct in assuming that these statements follow because the Sylow $p$-subgroups of $S_k$ must be cyclic of order $p$? They calculate the number of Sylow $p$-subgroups of $S_k$ by counting (number of $p$-cycles)/(number of $p$-cycles in a Sylow $p$-subgroup). They calculate the number of $p$-cycles in a Sylow $p$-subgroup to be $p(p-1)$, which I don't see.

-
You left out the important fact that $k=p$ or $p+1$. – Jonas Meyer May 22 '11 at 1:20
What do you mean, "the number of $p$-cycles in a Sylow $p$-subgroup"? – Arturo Magidin May 22 '11 at 2:34

If $k$ satisfies $p\leq k\lt 2p$, then $p$ is the largest power of $p$ that divides $k!$, so the $p$-Sylow subgroups of $S_k$ have order $p$.

If $\sigma\in S_k$ is written as a product of disjoint cycles, then the order of $\sigma$ is the least common multiple of the lengths of these cycles. In particular, an element has order a power of a prime $p$ if and only if it is a product of cycles whose lengths are powers of $p$. An element has order $p$ if and only if it is a product of (at least one) disjoint $p$-cycles.

Since $p\leq k\lt 2p$, the only elements of $S_k$ that are disjoint products of $p$-cycles are single $p$-cycles (not enough points to get another, disjoint, $p$-cycle). So a $p$-Sylow subgroup of such an $S_k$ will be cyclic of order $p$ (since $p|k!$ but $p^2\not|k!$), hence generated by an element of order $p$; and the only elements of order $p$ are the $p$-cycles. Conversely, any $p$-cycle generates a subgroup of order $p$, which will therefore be a $p$-Sylow subgroup of $S_k$. So the $p$-Sylow subgroups of $S_k$ are precisely the subgroups generated by $p$-cycles.

Any two distinct $p$-Sylow subgroups, $P_1$ and $P_2$, must intersect at proper subgroups of $P_1$ and $P_2$; but $P_i$ is of order $p$, so the only proper subgroup is the trivial subgroup, so $P_1\cap P_2 = \{1\}$.

I'm not sure what your final paragraph means, and I don't have Dummit and Foote with me right now, so I'll leave it for now.

-

Regarding the last part, I think you have written down the number of $p$-cycles in a fixed Sylow $p$-subgroup incorrectly ( or Dummit and Foote have a typo). The number of $p$-cycles in $S_p$ is $(p-1)!.$ To see this note that there are $p$ choices for the first entry, $p-1$ for the second, and so. This LOOKS like $p!$ choices, but notice that each $p$-cycle gets counted $p$ times this way. Each non-identity element of a given Sylow $p$-subgroup is a $p$-cycle, and there are $p-1$ non-identity elements of order $p$ in a group of prime order $p.$ Hence there are $(p-2)!$ different Sylow $p$-subgroups on $S_p.$ This implies that the normalizer of a Sylow $p$-subgroup of $S_p$ has order $p(p-1),$ which can also be seen directly. The argument for $S_{p+1}$ is of much the same nature.

-