Sylow $p$-subgroups in $S_p$ and order $p$ elements

Question

I realise this question has been asked a few times before, but I don't understand the answers.

What is the number of Sylow p subgroups in S_p?

• Why are the order $p$ elements in $S_p$ exactly cycles of the form $(1,a_2,a_3,\ldots,a_p)$? Sure that has order $p$, but why are all order $p$ elements on that form?
• Order $p$ subgroups of $S_p$ contain $p-1$ elements of order $p$, sure. Then Derek Holt says that the intersection of any two such subgroups is trivial. Why is that true?

Answer 1

For your first question: Let $\sigma \in S_p$. Then $\sigma$ can uniquely be written as a product of disjoint cycles. The concatenation of the lengths of these cycles is called the cycle-type of $\sigma$, for example, the cycle type of $(12)(34)$ in $S_5$ is $2$-$2$-$1$. Let $m_1$-..-$m_k$ be the cycle type of $\sigma$. Then $\text{order}(\sigma)=\text{lcm} (m_1,..,m_k)$. This can be seen by noticing disjoint cycles commute, so $\sigma^k=\text{id}$ if and only if $k$ is a multiple of $m_j$ for all $j$. Now suppose $\text{order}(\sigma)=p$, then $p=\text{lcm}(m_1,..,m_k)$ where $m_1$-..-$m_k$ is the cycle type of $\sigma$. Since $p$ is prime, $p$ is only a multiple of $1$ and $p$, so all $m_j$ are either $1$ or $p$. Obviously they cannot be all $1$ so one must be $p$, but then $k=1$ since $\sum_{i=1}^k m_i = p$. So the cycle type of $\sigma$ is $p$ and the result follows.

For your second question: suppose $H$ and $K$ are subgroups of $S_p$ of order $p$, $H \neq K$. Then $H \cap K$ is a subgroup of $H$, so its order is a divisor of $p$. If its order is $p$ then $H \cap K = H = K$, so its order must be $1$. But this implies $H \cap K$ is the trivial group.

Answer 2

First point: a cycle in $S_p$ is usually denoted with the smallest element encountered in the cycle as first element. As it is a cycle of length $p$, the smallest element is $1$.

For the second point, in any group, two subgroups of prime orders intersect trivially because of Lagrange's theorem: the order of the intersection divides the g.c.d. of the orders, so it is $1$ if the orders are different, and if the orders are the same, but the subgroups different.

Answer 3

On your second bullet: If the intersection is not trivial then it contains an element that has order $p$. This element generates both subgroups so they will coincide. Derek was dealing however with distinct subgroups.