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Trying to prove that A5 has no subgroups of order 20

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to prove that the alternating group on five letters, A5, contains no subgroups of order 20.


    2. Relevant equations
    I guess nothing is needed here, for this problem. Though I will use this extra space to explain my notation, if it would happen to differ from the standard one:

    Sym(5) - The symmetric group on five letters.
    A5 - The alternating group on five letters (The subgroup of Sym(5) consisting of all even permutations).


    3. The attempt at a solution
    I've tried to solve it for some time now, but none of my attemps have been fruitful at all. My first idea was to prove that the subgroup H of Sym(5) generated by a 5-cycle and a 4-cycle was, at least up to isomorphism, the only subgroup of Sym(5) having order 20. As the 4-cycle can be written as a product of 3 transpositions, it would then follow that the cycle is an odd permutation, and hence not in A5. However, I seem to be unable to prove this result.

    Another intuition I had was that, maybe, if d divides the order of a subgroup H of Sym(n) and d is smaller than or equal to n, then H must contain all d-cycles. If that were to be true then any such subgroup of order 20 would have to contain all 4-cycles, which are odd. Though, this too I seem to be unable to prove. As a matter of fact, I have no longer any idea if it is true or not.

    Anyway, I would appreciate some help with this, if possible.

    Thanks in advance
     
  2. jcsd
  3. Oct 10, 2009 #2

    Office_Shredder

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    Imagine I just pick a 5 cycle and I take the subgroup generated by it. This has 5 elements in it, but clearly does not contain all 5-cycles.

    Can you use Cauchy's Theorem for this? I think it gives you a good starting point (haven't worked it out all the way yet)
     
  4. Oct 10, 2009 #3
    I am sorry for that mistake. Figured it out when thinking of subgroups generated by a double transposition like (1,2)(2,3). It is indeed silly when thinking about it.

    However, I would like to avoid using Cauchy's Theorem, as it is presented as an excercise far earlier than the theorem in the material in which I found the problem. What about my first idea? It seems like the only possible subgroups of order 20 of Sym(5) are the ones I mentioned, and they are certainly not contained in A5, as any 4-cycle is odd. Have any ideas on how to investigate this further?

    Thanks for your reply!
     
  5. Oct 11, 2009 #4
    I came up with something that seems to be proving the assertion. Though, I would not say that I am sure that everything in it is correct. Could someone please read through and comment on it?

    Here it goes: First I define a concept that I use in the proof. I have no idea if what I define is consistent with the usual definition of cycle structure, but it works for the purpouse of this proof.

    Definition: If [tex]\sigma \in S{n}[/tex] we know that we can write it as a product of disjoint cycles. We characterize [tex]\sigma[/tex] in the following way;
    If [tex]\sigma[/tex] is a product of [tex]\tau_{k_{1}}, \ldots \tau_{k_{r}}[/tex], where the [tex]\tau_{k_{i}}[/tex] are disjoint cycles of length [tex]k_{i}[/tex], we say that the cycle structure of [tex]\sigma[/tex] is [tex]\sum{k_{i}}[/tex], denoted [tex]\mathcal{S}(\sigma) = \sum{k_{i}}[/tex].

    Proof: Assume [tex]H \leq A_{5}[/tex], [tex]|H| = 20[/tex]. If [tex]\sigma \in H[/tex] then [tex]\mathcal{S}(\sigma) \in \{\phi,\: 2+2,\: 5\}[/tex], as if the structure would have been 2,4,3+2 it would have been an odd permutation. As the order of a 3-cycle is 3, we get by Lagrange's Theorem that no 3-cycle can be an element of a group of order 20. Thus we know that any element [tex]\sigma \in H[/tex] must be either a product of two disjoint transpositions, a 5-cycle or a product of these. We now examine the subgroups of [tex]A_{5}[/tex] having only elements of said type.

    Let [tex]\tau[/tex] and [tex]\sigma[/tex] be of structure 2+2 and 5, respectively. Consider the subgroup [tex]K = \langle(\tau, \sigma)\rangle[/tex]. As there exists 5 distinct powers of [tex]\sigma[/tex] and 2 distinct powers of [tex]\tau[/tex] (conclusion drawn by the orders of the subgroups generated by one of the elements), we know that [tex]|K| \geq 5+2-1[/tex], as the identity is counted twice. Also, as the [tex]\tau\sigma^{k}[/tex] are distinct for [tex]k = 1,2,3,4,5[/tex], and only for [tex]k = 5[/tex] we have [tex]\tau\sigma^{k} = \tau[/tex], we get [tex]|K| \geq 10[/tex]. If [tex]\tau[/tex] conjugates [tex]\sigma[/tex] to a power of itself, equality holds. Even if one would not guess that this tells us anything, it actually reduces the problem quite alot.

    If we are to arrive at a subgroup of order 20 we must adjoin some element of the appropriate type to the generating set. However, if we adjoin a 5-cycle [tex]\sigma'[/tex], we have that, if [tex]K' = \langle(\tau, \sigma', \sigma)\rangle[/tex], [tex]|K'| \geq 40[/tex], so it will generate the whole of [tex]A_{5}[/tex]. If we adjoin another double-transposition [tex]\tau'[/tex] perhaps? Well, if [tex]K'' = \langle(\tau, \tau', \sigma)\rangle[/tex], for every [tex]\rho \in K[/tex] we obtain [tex]\tau'\rho \in K'' \backslash K[/tex], and they are by the cancellation laws with respect to [tex]\tau'[/tex] all distinct. Thus [tex]|K''| \geq 20[/tex]. Can we have equality? Well, no, as [tex]\tau[/tex] and [tex]\sigma[/tex] does not commute. This is due to the fact that, if they would commute, so [tex]\tau\sigma = \sigma\tau[/tex], then [tex]\tau\sigma\tau = \tau^{2}\sigma = \sigma[/tex]. But if [tex]\sigma = (a_1, a_2, a_3, a_4, a_5)[/tex] and [tex]\tau = (a_1, a_k)\tau_x[/tex] where [tex]\tau_x[/tex] is a transposition (we can always modify [tex]\sigma[/tex] to begin with the same letter as [tex]\tau[/tex]), then [tex]\tau\sigma\tau[/tex] moves [tex]a_1[/tex] to [tex]a_k[/tex], further to [tex]a_{k+1}[/tex] and we would be forced to require [tex]\tau[/tex] to move [tex]a_k+1[/tex] back to [tex]a_{2}[/tex]. Thus [tex]\tau = (a_1, a_k)(a_k + 1, a_2)[/tex], so [tex]\tau\sigma\tau[/tex] moves [tex]a_2[/tex] to [tex]a_{k+1}[/tex], and then to [tex]a_{k+2}[/tex]. As [tex]\sigma[/tex] moves [tex]a_2[/tex] to [tex]a_3[/tex], we would need [tex]k+2 = 3[/tex], so [tex]k=1[/tex]. But then [tex]\tau[/tex] would certainly not be a double transposition, which is a contradiction.

    Thus we have [tex]|K''| > 20[/tex]. As every step in the construction was necessary to even keep the possibility of [tex]K'' = H[/tex], the supposed subgroup of order 20 of [tex]A_5[/tex], we can be sure that no such subgroup exists.
     
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