4

Let $G$ be a subgroup of the symmetric group $S_n$.

Suppose that $G$ has the following property:

given any positive integer $k\leq n$, there exists an element of $G$ having a cycle of length exactly $k$.

Is it any true that under this assumption $G=S_n$? If not, could one make a counterexample with $n$ prime?

Please notice that I am not saying that $G$ has elements which ARE cycles of order $k$.

Could you please remove the duplicate, as if you read it carefully you see that it is a different question? (please see also the answer that I accepted)

Reyx_0
  • 1,128
  • 2
    A $n$- cycle and a transposition generate $S_n$, so.. – Arkady Feb 21 '16 at 14:03
  • but that is not the situation, as the transposition may not occur in G – Reyx_0 Feb 21 '16 at 14:05
  • Your $G$ contains an order 2 element which is also called a transposition..Wait, when you say an element having a cycle of length $k$, I interpreted it to mean that the element IS a cycle of length $k$. Is this not the case? – Arkady Feb 21 '16 at 14:07
  • I mean "an element having a cycle of length k", not "a cycle of length k". :) – Reyx_0 Feb 21 '16 at 14:08
  • What does it mean to have a cycle of length $k$? Do you mean in a disjoint cycle decomposition? – Arkady Feb 21 '16 at 14:10
  • exactly: any element can be decomposed in cycles. the property is that for any k, you can find an element in G such that when you decompose it in cycles has one of order k. In other words, you can always find an element with an orbit of size k (when you consider the obvious action on 1... n). – Reyx_0 Feb 21 '16 at 14:13
  • 1
    @Jake, but it is not true that an arbitrary transposition and an arbitrary $n$-cycle generate $S_{n}$. – Andreas Caranti Feb 21 '16 at 14:15
  • There are no counterexamples for $n$ prime, as it is well-known that in this case $S_{n}$ is generated by a $2$-cycle and an $n$-cycle.

    In the general case, this is not that obvious, as for instance $S_{4}$ is not generated by $(12)$ and $(1324)$.

     – Andreas Caranti Feb 21 '16 at 14:15
  • This question is a special case of the question I mentioned as a possible duplicate. Note that this answer---http://math.stackexchange.com/a/1202285/155629---implies that that the answer to the question here is yes, as any transitive subgroup of $S_n$ containing a $2$-cycle and an $(n - 1)$-cycle is transitive. – Travis Willse Feb 21 '16 at 14:20
  • @Travis did you see my edit on the question? why G should contain a 2 cycle? – Reyx_0 Feb 21 '16 at 14:23
  • I only saw the edit now. By "length of cycle of an element" do you just mean the order of the element? – Travis Willse Feb 21 '16 at 14:37
  • (If so, the answer is no: In $S_6$ the elements $1, (12), (123), (1234), (12345), (12)(345)$ have respective orders $1, 2, 3, 4, 5, 6$, but the subgroup they generate fixes $6$ and so isn't even transitive.) – Travis Willse Feb 21 '16 at 14:39
  • no, I mean the orbit of an element in {1...n} after acting with the powers of the permutation, please see the comments above :) – Reyx_0 Feb 21 '16 at 14:42
  • @Travis for examples there might be and element of the form (12)(345) and this gives cycles of length 2 and 3 (not one of order 6) – Reyx_0 Feb 21 '16 at 14:51
  • Ah, I see; in that case you might edit the original question to reflect this, the wording as it is reads a bit deceptively to me. – Travis Willse Feb 21 '16 at 14:54
  • Actually I am not completely sure how to rephrase it without making it too pedantic... if you have an idea please modify it! I am sure it is deceptive, as it also reads very similiarly to other kind of questions (as all the comments have shown...!). – Reyx_0 Feb 21 '16 at 15:00

1 Answers1

3

The answer is yes. There is a result (I think of Jordan) that a transitive subgroup of $S_n$ that contains an element of prime order $p$ with $n/2 < p < n-2$ contains $A_n$. (This result is used in a probabilistic algorithm for testing whether a given group contains $A_n$.)

For $n \ge 8$ there exists such a prime $p$. By assumption, your group contains an $n$-cycle, so it is transitive, and a $p$-cycle, so it contains $A_n$. (The assumptions say only that $G$ contains an element containing a $p$-cycle, but since $p>n/2$, some power of that element must be a $p$-cycle.) Since it contains an $n$-cycle and an $(n-1)$-cycle, it must contain an odd permutation, and hence it must be $S_n$.

You can check the cases $n\le 7$ individually.

References for the result of Jordan are:

C. Jordan. Sur la limite de transitivit\'e des groupes non altern\'es. Bull. Soc. Math. France, 1:40-71, 1873.

Theorem 13.9 of Wielandt's book|"Finite Permutation groups".

(The result actually applies to primitive groups with any prime $p<n-2$. Ths condition $p>n/2$ ensures that the group is primitive.)

The existence of the prime follows from a strong version of Bertrand's postulate - see the Wikipedia page for example.

Derek Holt
  • 90,008