1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook