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    I've got my final exam for Group Theory tomorrow and I just need a little spot of help with a few things, hopefully you'll be able to help

    1) Say we have three permutations (in S_5) in disjoint cycle form:

     p = (123)(45)
     q = (24)(15)
     r = (143)(25)

    We know from cycle structure that p and r are conjugate but q is not conjugate to either. But then how do you go about calculating the congujating element?
    I know that it is of the form x^{-1}px = r, but I don't know how to actually calculate it.

    Also, how would you write each one as a product of transpositions? (I know that q doesn't change in this respect, but I wonder whether how I normally calculate it is the correct way)

    2) How would you find the largest possible order of an element in a given non-cyclic group, say A_5?

    3) Prove (using Sylow's Theorem) that there is no simple group of order 56.

    With this one I know how to start it, by saying that 56 = 2^3 . 7, so we look at Sylow-2-subgroups and Sylow-7-subgroups.

    We then have n_2|7 and n_2 \equiv 1 mod2

    and similarly n_7|8 and n_7 \equiv 1 mod7.

    From this we see that n_7 = 8 and n_2 = 1 or 7

    I imagine the next stage would be to assume that G is simple such that n_2 = 7, but I don't understand how I would go about showing that this leads to a contradiction/impossibility.

    Thanks a lot if you are able to help. :yes:
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    For 1, I believe the general method is something like this:
    say P is (A1, A2, ..., Am)...(Z1, Z2, ..., Zn) and p is (a1, a2, ..., am)...(z1, z2, ..., zn), both written as a product of disjoint cycles.

    Then x st A1->a1, A2->a2, ...., Z1->z1, ..., Zn->zn satisfies x^(-1) P x = p (assuming composition from left to right).
    This makes sense. Eg, on the LHS, x^(-1) takes a1 to A1, P takes A1 to A2, and x takes A2 to a2. On the RHS, p takes a1 to a2.

    For 3, you missed n7=1. If either of n2 or n7 is 1, G is not simple. Suppose n7 is not 1. So n7=8. Each Sylow 7-subgroup has size 7 and any two of them can only intersect trivially. So there are 8*(7-1)=48 elements of order 7 in G. That leaves 8 other elements. These must make up a single Sylow 2-subgroup so n2=1.
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    (Original post by marcusmerehay)
     p = (123)(45)

    Also, how would you write each one as a product of transpositions? (I know that q doesn't change in this respect, but I wonder whether how I normally calculate it is the correct way)
    I believe (123)(45) = (13)(12)(45)
    and (143)(25) = (14)(13)(25)
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    (Original post by Calira)
    I believe (123)(45) = (13)(12)(45)
    and (143)(25) = (14)(13)(25)
    That's what I thought.
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    (Original post by SsEe)
    For 1, I believe the general method is something like this:
    say P is (A1, A2, ..., Am)...(Z1, Z2, ..., Zn) and p is (a1, a2, ..., am)...(z1, z2, ..., zn), both written as a product of disjoint cycles.

    Then x st A1->a1, A2->a2, ...., Z1->z1, ..., Zn->zn satisfies x^(-1) P x = p (assuming composition from left to right).
    This makes sense. Eg, on the LHS, x^(-1) takes a1 to A1, P takes A1 to A2, and x takes A2 to a2. On the RHS, p takes a1 to a2.

    For 3, you missed n7=1. If either of n2 or n7 is 1, G is not simple. Suppose n7 is not 1. So n7=8. Each Sylow 7-subgroup has size 7 and any two of them can only intersect trivially. So there are 8*(7-1)=48 elements of order 7 in G. That leaves 8 other elements. These must make up a single Sylow 2-subgroup so n2=1.
    Thanks for the help on the Sylow question, much appreciated. Any chance you could show me how you would work through the method for 1 with my p and r?
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    (Original post by marcusmerehay)
    Thanks for the help on the Sylow question, much appreciated. Any chance you could show me how you would work through the method for 1 with my p and r?
    p = (1, 2, 3)(4, 5)
    r = (1, 4, 3)(2, 5)

    So (think of it like the other way of writing a permutation), x maps 1->1, 2->4, 3->3, 4->2, 5->5. In other words x = (2, 4). Make sure you see why doing this works.

    Also, a cycle (a1, a2, ..., an) as a product of transpositions is just:
    (a1, a2)(a1, a3)(a1, a4)...(a1, an)

    For question 2, I don't know any general approach and a quick google search suggests that nobody else has come up with one either. But in special cases like the symmetric/alternating groups, you know that if you write an element, p, as a product of disjoint cycles of lengths L1, L2, ..., Ln then the order of p is lcm(L1, ..., Ln). So one approach is just to list all the possible cycle types and work out the highest order. So for S_4 you've got 4, 3-1, 2-2, 2-1-1 and 1-1-1-1. The highest order is 4, possessed by elements of cylce type 4. For A_5, you've got 5, 3-1-1, 2-2-1 and 1-1-1-1-1 so the highest order is 5.
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    (Original post by SsEe)
    p = (1, 2, 3)(4, 5)
    r = (1, 4, 3)(2, 5)

    So (think of it like the other way of writing a permutation), x maps 1->1, 2->4, 3->3, 4->2, 5->5. In other words x = (2, 4). Make sure you see why doing this works.

    Also, a cycle (a1, a2, ..., an) as a product of transpositions is just:
    (a1, a2)(a1, a3)(a1, a4)...(a1, an)

    For question 2, I don't know any general approach and a quick google search suggests that nobody else has come up with one either. But in special cases like the symmetric/alternating groups, you know that if you write an element, p, as a product of disjoint cycles of lengths L1, L2, ..., Ln then the order of p is lcm(L1, ..., Ln). So one approach is just to list all the possible cycle types and work out the highest order. So for S_4 you've got 4, 3-1, 2-2, 2-1-1 and 1-1-1-1. The highest order is 4, possessed by elements of cylce type 4 or 2-2. For A_5, you've got 5, 3-1-1, 2-2-1 and 1-1-1-1-1 so the highest order is 5.
    Strangely enough I'd just managed to get the thing about the alternating group before you replied. Thank you very much for explaining the first question, that actually makes sense now! :p:
 
 
 
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