Find the number of elements of each order in a simple group of order $60$ Question: Find the number of elements of each order in a simple group of order $60$ without recourse to the fact that the group is isomorphic to the alternating group $A_5$.
Work: So I did some work already and was able to find the number of elements of orders 3 and 5 by finding the number of Sylow 3- and 5-subgroups. This has me at 1 element of order 1, 20 of order 3 and 24 of order 5. However, I am lost as to how to do the rest. Indeed, there seem to be two key issues that I'm missing:
1) I can deduce $n_2 \in \{5,15\}$, where $n_2$ is the number of 2-Sylow subgroups. However, I cannot fully establish that $n_2 = 5$. I was thinking of a contradiction, assuming that $n_2 = 15$ and then showing that since the intersection of the subgroups has order at most 2, it then follows that there are more than 15 distinct elements in total in Sylow 2-subgroups, a contradiction. I'm not sure I can justify this or if it is even true necessarily from this argument.
2) Even if I do somehow handwave my way around part 1), the bigger issue is establishing that there are no elements of order 4, i.e. that all elements in 2-Sylow subgroups have order 2.
I've spent a considerable chunk of time on this already, so any help (perhaps more than merely a hint :)) would be appreciated.
 A: What you need to show is that for $H$ and $K$ different Sylow 2 subgroups, $H\cap K=<1>$.
Once this is established, there couldn't be 15 Sylow 2 subgroups -- we'd get too many elements together with your determination of numbers of Sylow 3 and Sylow 5 subgroups.  So there are 5 Sylow 2 subgroups.  Since H is not cyclic, there are exactly 3 elements of order 2 in H, for a total of 15 in G.
Back to $H\cap K=<1>$.  Suppose $H\cap K\neq <1>$.  Then $H\cap K=<x>$ where $x$ is of order 2.  Then $C_G(x)$ has order greater than 4.  Use Sylow to argue that this is impossible.
 Holt has already answered why a Sylow 2 subgroup can not be cyclic.  Here's an expansion of his answer:
 In any permutation group if the cycle $c=(i_1,i_2,\cdots ,i_n)$ has length n, the parity of $c$ is $(-1)^{n-1}$.  So $c$ is an odd permutation (not a member of the alternating group) if and only if $n$ is even.  Furthermore, the parity of a product is the product of the parities.
 Suppose G is a group and a Sylow 2 subgroup of G is cyclic of order 4, say the order of G is 4m with m odd.  Then G is not simple:
  Consider G as a subgroup of the symmetric group on the elements of G (Cayley representation).  Let $g\in G$ be of order 4 and $x\in G$.  Then the cycle of $g$ containing $x$ is $(x,xg,xg^2,xg^3)$.  Thus the cyclic decomposition of $g$ consists of m 4 cycles.  By the above paragraph, the parity of $g$ is $((-1)^3)^m=-1$.  So $g\notin A_{4m}$.  So $S_{4m}=GA_{4m}$ ($A_{4m}$ has index 2).  Hence $$S_{4m}/A_{4m}\cong{A_{4m}\over A_{4m}\cap G}$$.  Then $A_{4m}\cap G$ is a normal subgroup of G of index 2.
