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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?
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JACK3D
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3 Answers3

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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.

M. Van
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    Great answer thank you! This makes everything clear – JACK3D Jul 24 '16 at 13:27
  • I do not understand how $k=1$, could you explain? Why $\sum_{i=1}^km_i=p?$ Thank you in advance. – xyz Sep 01 '20 at 07:27
  • Why $k$ must be 1? as one of $m_k$ can be $1$ as one of the cycle can be identity. – xyz Sep 01 '20 at 13:53
  • any $\sigma \in S_p$ can be written as $\sigma=\tau_1 .. \tau_k$ where the $\tau_j$ are disjoint cycles of length $m_j$ and $\sum m_j = p$ (any number between $1$ and $p$ will appear exactly once in this expression). If all $\m_j$ are $1$ then $\sigma$ is the identity. If not, then at least one $\m_j$ is $p$. But since this is a sum of natural numbers $\geq 1$ summing to $p$, one of them being equal to $p$ implies there can’t be any others, so $k=1$. Clear? – M. Van Sep 08 '20 at 11:31
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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.

darij grinberg
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Bernard
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  • The point is why $p$ must be a cycle of length $p$. Why? I do not understand. Thank you in advance. – xyz Sep 01 '20 at 13:54
  • @briantylf: There seems to be some misunderstanding – $p$ is not a cycle, it is the length of a cycle in $S_p$. – Bernard Sep 01 '20 at 14:20
  • Sorry. The question should be why a element $\sigma$ in $S_p$ with $|\sigma|=p$ must be a $p$ cycle? – xyz Sep 01 '20 at 14:24
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    That's because any permutation is a product of disjoint cycles, and its order is the l.c.m. of the orders of the components. If the l.c.m. is prime, you don't have many choices as to the decomposition. – Bernard Sep 01 '20 at 14:30
  • Yes. If $\sigma=\sigma_1...\sigma_n$, where $\sigma_1,...\sigma_n$ are disjoint cycles. Then $|\sigma|=lcm(|\sigma_1|,...,|\sigma_n|)$. Then $|\sigma_i|$ can be either $1$ or $p$. Since they are disjoint, I can take $|\sigma_1|=1$, then $\sigma_1=id$. However, as $\sigma_2,...,\sigma_n$ are disjoint $p$ cycles, how to make sure their composition $\sigma$ is still a cycle? – xyz Sep 01 '20 at 14:45
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    There can't be two disjoint $p$-cycles since we're in S_p$. – Bernard Sep 01 '20 at 14:47
  • oh yes. Thank you very much. – xyz Sep 01 '20 at 14:51
  • Do you have recommended reference for symmetric groups? I have found many books in abstract algebra, but none of them provide good reference for symmetric group and cycles – xyz Sep 01 '20 at 14:53
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    Not really. As may people from my generation, I learnt with Bourbaki, but it's not really recommended for beginners. I know another general book at the introductory level, which is well written, but it's in french. – Bernard Sep 01 '20 at 16:00
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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.

darij grinberg
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drhab
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