How many subgroups of size 5 in A_6 have cyclic elements?

[alex07966]

I need to find the number of subgroups of size 5 in A_6.


I have started by noting that as the subgroup size is 5, a prime, the subgroups must be cyclic. I have worked out that there are 144 elements of order 5 in A_6, but this can't be equal to the number of subgroups (i found two subgroups which have the same elements in!). Someone please help!

 

[Dick]

If two subgroups of order 5 intersect, can you describe the intersection?

 

[alex07966]

Ok, I have noticed that <a> = <a^2> = <a^3> = <a^4> for all a in A_6 where a is a 5-cycle. So this means that the number of elements of order 5 must be divided by 4. Hence the answer is 144/4 = 36 subgroups of size 5 in A_6.
Is this correct?

 

[Dick]

Ok, I have noticed that <a> = <a^2> = <a^3> = <a^4> for all a in A_6 where a is a 5-cycle. So this means that the number of elements of order 5 must be divided by 4. Hence the answer is 144/4 = 36 subgroups of size 5 in A_6.
Is this correct?

Yes, every subgroup of order 5 contains four of them and no element of order 5 is contained in two different subgroups.