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##S_p## generated by any transposition and p-cycle

  1. May 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##p## be prime. Show that ##S_p = \langle \tau, \sigma##, where ##\tau## is any transposition and ##\sigma## any ##p##-cycle.

    2. Relevant equations


    3. The attempt at a solution

    I read somewhere that it suffices to prove this for ##\sigma = (1,2,...,p)##. Intuitively, this is clear, but I want to justify this. Let ##\sigma = (x_1,.x_2,...,x_p)##. Let ##f : \{1,...,p\} \to \{1,...,p\}## be defined ##f(x_i) = i## and fixes everything else. It seems that we need to show that ##S_p = \langle \tau, \sigma \rangle## if and only if ##S_p = \langle f \tau, f \sigma \rangle##, which can be proven by showing ##\langle f \tau , f \sigma \rangle = f \langle \tau, \sigma \rangle##, in order to justify this reduction.

    Showing ##\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle## is rather simple, but I am having trouble with the other inclusion. I am probably overlooking some trivial fact. I could use a hint.
     
  2. jcsd
  3. May 26, 2017 #2

    fresh_42

    Staff: Mentor

    What is ##\langle f\tau,f\sigma \rangle## after you managed to prove the result for ##(1,\ldots ,p)\,##?
     
  4. May 26, 2017 #3
    I was thinking that ##\langle gx,gy \rangle = \langle g \langle x,y \rangle## was a result that held for all groups. Is that not true? I should have specified that I would like to prove that it is true for all groups.
     
  5. May 26, 2017 #4

    fresh_42

    Staff: Mentor

    This is not what I meant. Far simpler than this. I asked basically: What are ##f \tau## and ##f \sigma##, what do they look like? And then, if you managed to prove the result in the special case of transposition and ##p-##cycle ##(1,\ldots ,p)##, what is ##\langle f\tau , f \sigma \rangle## for a subgroup?
     
  6. May 26, 2017 #5
    Well, ##f \tau## would just be some transposition and ##f \sigma## would give us back ##(x_1,...,x_p)##, right?
     
  7. May 26, 2017 #6

    fresh_42

    Staff: Mentor

    Yes. Since we have no further condition on the transposition used, we can use ##\tau## as well as ##f \tau##. It makes no difference for "any".
    O.k., beside of the part "fixes everything else" which I don't see why you want to have this, yourself has defined ##\sigma = (x_1,x_2, \ldots,x_p)## so I assumed ##f \sigma = (1, \ldots ,p)##. I thought this was the formal notation for renumbering the elements, which we actually want to do. Now assume we have proven, that ##S_p## is generated by some transposition ##f \tau ## and the ##p-##cycle ##f \sigma = (1,\ldots ,p)##. This makes ##\langle f\tau , f \sigma \rangle = S_p##. You further have said
    which leaves us with the situation ##S_p=\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle## rather simple and I simply wanted to point you to
    $$
    f \langle \tau, \sigma \rangle \subseteq S_p = \langle f \tau , f \sigma \rangle
    $$ is by far even simpler!
     
  8. May 29, 2017 #7
    I think there might be a problem. ##f \langle \tau, \sigma## is a left-coset and therefore isn't generally a group. So, even though ##f \langle \tau, \sigma \rangle## contains the generators ##f \tau## and ##f \sigma##, this doesn't necessarily imply that it contains ##\langle f \tau, f \sigma##.

    At this point, I am not sure how to fix the proof.
     
  9. Jun 6, 2017 #8

    fresh_42

    Staff: Mentor

    I'm not sure what you mean. ##f\langle \tau, \sigma \rangle = \{fg\,\vert \,g \in \langle \tau, \sigma \rangle\}##. I haven't checked your "rather simple" part, but as ##S_p## is the entire group, it contains of course its elements ##fg##.
     
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