# 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 group with a cyclic Sylow $2$-subgroup cannot be simple (so there are no elements of order $4$). To see that, consider the image of the regular representation of $G$ as a subgroup of $S_{60}$. The image of an element of order $4$ would be an odd permutation, which contradicts simplicity. – Derek Holt Dec 2 '14 at 2:04
• @DerekHolt, can you be more explicit? I don't really follow. I haven't learned about regular representations, so how would I consider this as a subgroup of $S_{60}$? Is it just multiplication on the left? But even if so, why would it be an odd permutation? Ugh, and I also don't see the connection between odd permutations and simplicity (or normal groups). – Ryker Dec 2 '14 at 2:35
• I am surprised that you are familiar with Sylow's Theorems but have never come across Cayley's Theorem, which is usually one of the first results in group theory. Yes, it is just multiplication on the left. It's an odd permutation because it is 15 $4$-cycles. Intersecting with the alternating group gives a normal subgroup of index $2$, contradicting simplicity. – Derek Holt Dec 2 '14 at 2:50
• @DerekHolt, I was aware of Cayley's Theorem, I just wasn't sure what exactly you were referring to with the image of the regular representation. But why would the image be 15 $4$-cycles? And when you say "intersecting with the alternating group", intersecting what with the alternating group? The representation of $G$? If so, I don't see why the intersection of this subgroup of $S_{60}$ would necessarily have to have index $2$. I assume the odd permutation has to come into play somehow, but how? I mean, it's just one of the elements. – Ryker Dec 2 '14 at 3:57
• You could see this post for example: math.stackexchange.com/questions/919119 – Derek Holt Dec 2 '14 at 4:40

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.
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.
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.
• You say "Since $H$ is not cyclic", but I think that's the hardest thing to show, and that's what my question was about. I.e. how do I show that? – Ryker Dec 2 '14 at 4:00