Find Sylow subgroups \(P_2, P_3\), and \(P_5\) for the three primes \(2, 3, 5\) dividing the order of \(A_5\). (It suffices to find generators of each of these groups.)
Find the normalizers in \(A_5\) of each of the three groups you found in part (a).
For each of the divisors \(d = 6, 10, 15\) of 60, either find a subgroup of \(A_5\) of order \(d\) or explain why none exists for that choice of \(d\).
Solution
Since \(\abs{A_5}=5!/2=60=2^2\cdot3\cdot5\), Sylow \(2\)-, \(3\)-, and \(5\)-subgroups of \(A_5\) have orders \(4\), \(3\), and \(5\).
1. For part (a), take\[ P_2=\langle(1\,2)(3\,4),(1\,3)(2\,4)\rangle,\qquad P_3=\langle(1\,2\,3)\rangle,\qquad P_5=\langle(1\,2\,3\,4\,5)\rangle. \]
The double transpositions \((1\,2)(3\,4)\), \((1\,3)(2\,4)\), \((1\,4)(2\,3)\) are even, commute, and together with \(1\) form the Klein four-group \(V_4\) of order \(4\). A \(3\)-cycle and a \(5\)-cycle are even and generate cyclic subgroups of orders \(3\) and \(5\).
2. For part (b),\[ N_{A_5}(P_2)=A_4,\qquad N_{A_5}(P_3)=\langle(1\,2\,3),(1\,2)(4\,5)\rangle\cong S_3,\qquad N_{A_5}(P_5)=\langle(1\,2\,3\,4\,5),(2\,5)(3\,4)\rangle\cong D_{10}, \]where \(A_4\) is the alternating group on \(\{1,2,3,4\}\) and \(D_{10}\) is dihedral of order \(10\).
The index of the normalizer of a Sylow subgroup is the number of Sylow subgroups for that prime [@DF04], so it suffices to count Sylow subgroups and exhibit a subgroup of the right order inside each normalizer. The \(15\) double transpositions of \(A_5\) lie three to a Klein four-group, so \(n_2=5\) and \(\abs{N_{A_5}(P_2)}=12\); since \(V_4\lhd A_4\), the normalizer is \(A_4\). The \(3\)-cycles in \(A_5\) number \(20\), so \(n_3=10\) and \(\abs{N_{A_5}(P_3)}=6\); conjugation by \((1\,2)(4\,5)\) sends \((1\,2\,3)\) to \((2\,1\,3)=(1\,2\,3)^{-1}\), so the displayed group of order \(6\) normalizes \(P_3\), and it is nonabelian, hence \(\cong S_3\). The \(5\)-cycles number \(24\), so \(n_5=6\) and \(\abs{N_{A_5}(P_5)}=10\); conjugation by \((2\,5)(3\,4)\) sends \((1\,2\,3\,4\,5)\) to \((1\,5\,4\,3\,2)=(1\,2\,3\,4\,5)^{-1}\), so the displayed dihedral group of order \(10\) is the normalizer.
3. For part (c), \(A_5\) has subgroups of orders \(6\) and \(10\) and none of order \(15\).
The normalizers of \(P_3\) and \(P_5\) in step 2 have orders \(6\) and \(10\). A group of order \(15\) is cyclic, because its Sylow \(3\)- and \(5\)-subgroups are both unique and hence normal, so it is their direct product \(\ZZ/3\times\ZZ/5\cong\ZZ/15\). The elements of \(A_5\) have cycle types \(1^5\), \(2^21\), \(31^2\), \(5\), hence orders \(1,2,3,5\); none has order \(15\).
4. Q.E.D.
Steps 1, 2, and 3 answer parts (a), (b), and (c).