# Trying to prove that A5 has no subgroups of order 20

1. Oct 10, 2009

### Jösus

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.

2. Oct 10, 2009

### Office_Shredder

Staff Emeritus
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)

3. Oct 10, 2009

### Jösus

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?

4. Oct 11, 2009

### Jösus

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

Proof: Assume $$H \leq A_{5}$$, $$|H| = 20$$. If $$\sigma \in H$$ then $$\mathcal{S}(\sigma) \in \{\phi,\: 2+2,\: 5\}$$, 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 $$\sigma \in H$$ must be either a product of two disjoint transpositions, a 5-cycle or a product of these. We now examine the subgroups of $$A_{5}$$ having only elements of said type.

Let $$\tau$$ and $$\sigma$$ be of structure 2+2 and 5, respectively. Consider the subgroup $$K = \langle(\tau, \sigma)\rangle$$. As there exists 5 distinct powers of $$\sigma$$ and 2 distinct powers of $$\tau$$ (conclusion drawn by the orders of the subgroups generated by one of the elements), we know that $$|K| \geq 5+2-1$$, as the identity is counted twice. Also, as the $$\tau\sigma^{k}$$ are distinct for $$k = 1,2,3,4,5$$, and only for $$k = 5$$ we have $$\tau\sigma^{k} = \tau$$, we get $$|K| \geq 10$$. If $$\tau$$ conjugates $$\sigma$$ 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 $$\sigma'$$, we have that, if $$K' = \langle(\tau, \sigma', \sigma)\rangle$$, $$|K'| \geq 40$$, so it will generate the whole of $$A_{5}$$. If we adjoin another double-transposition $$\tau'$$ perhaps? Well, if $$K'' = \langle(\tau, \tau', \sigma)\rangle$$, for every $$\rho \in K$$ we obtain $$\tau'\rho \in K'' \backslash K$$, and they are by the cancellation laws with respect to $$\tau'$$ all distinct. Thus $$|K''| \geq 20$$. Can we have equality? Well, no, as $$\tau$$ and $$\sigma$$ does not commute. This is due to the fact that, if they would commute, so $$\tau\sigma = \sigma\tau$$, then $$\tau\sigma\tau = \tau^{2}\sigma = \sigma$$. But if $$\sigma = (a_1, a_2, a_3, a_4, a_5)$$ and $$\tau = (a_1, a_k)\tau_x$$ where $$\tau_x$$ is a transposition (we can always modify $$\sigma$$ to begin with the same letter as $$\tau$$), then $$\tau\sigma\tau$$ moves $$a_1$$ to $$a_k$$, further to $$a_{k+1}$$ and we would be forced to require $$\tau$$ to move $$a_k+1$$ back to $$a_{2}$$. Thus $$\tau = (a_1, a_k)(a_k + 1, a_2)$$, so $$\tau\sigma\tau$$ moves $$a_2$$ to $$a_{k+1}$$, and then to $$a_{k+2}$$. As $$\sigma$$ moves $$a_2$$ to $$a_3$$, we would need $$k+2 = 3$$, so $$k=1$$. But then $$\tau$$ would certainly not be a double transposition, which is a contradiction.

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