0

$H$ is a subgroup of $A_5$ that has order 30. From this I know that $|A_5 : H|$ = 2. From this I'm supposed to prove that $H$ contains all 3 and 5 cycles and then use that to prove that there cannot be a subgroup of order 30 in $A_5$. I understand how to do the 3-cycle, but I'm very confused on the 5 cycle. Is there anyway to prove that without using simplicity of $A_5$ or conjugacy classes?

Jay3
  • 307
  • Perhaps this will help: $H$ must have a subgroup of order 15. – Taylor Dec 02 '15 at 17:43
  • No- I'm not able to use simplicity, commutator subgroups or conjugacy classes in my answer. I can only prove that the cycles are elements of the subgroup and use counting techniques. – Jay3 Dec 02 '15 at 17:54
  • 1
    If $H$ contains all $3$-cycles, then $H$ must be all of $A_{5}$, since the $3$-cycles generate $A_{5}$. This is usually a step en route to the simplicity of $A_{5}$. – Alex Wertheim Dec 02 '15 at 18:14

0 Answers0