How to find all elements of S4 that satisfy the equation x^4=e?

[AdrianZ]

Okay, you counted it right, but then, you already knew it should be 3.

Anyway, I believe you were going to set up a generalized formula for k-cycles in Sn.
When you have that, you may want to revisit this problem.

Actually I wanted to set up a generalized formula for the number of solutions of the equation xⁿ=e, but now I see that it can be a little bit more tricky when n is not a prime number, because then I'll have to count the number of the generated cycles as products of disjoint cycles. that would make it harder.

The number of k-cycles in S_n is (n,k)*(k-1)! as someone else mentioned. the reason is that first we have to choose k letters out of n letters for forming k-cycles, then we fix the first element and permute the others and that can be done in (k-1)! ways. so the answer will be (n,k)*(k-1)! where (n,k) is n choose k.

this gives us the ability to predict the solutions of xⁿ=e when n is prime. since the only divisors of n are 1 and itself, we'll have (1 + (n-1)!) solutions.
so, x⁵=e we'll have 25 solutions in S₅.

we can also predict the number of solutions of x^k=e in S_n when n is prime. the answer will be (1 + (n,k)*(k-1)!).
the number of solutions of x⁴=e in S₅ is 1 + 5*24 = 121.
the number of solutions of x²=e in S₃ is 1 + 3*1 = 4. those solutions namely are: {e,(1 2),(1 3),(2 3)}.

the case where n is not prime is a bit tricky, but I'll think about it. first I'll need to prove some theorems, for example if p and q are two disjoint cycles, then o(pq)=o(p)o(q). it needs more considerations, I'll think about it later.

Thanks guys for the help, and have a nice thanksgiving holiday tomorrow.

 

[I like Serena]

this gives us the ability to predict the solutions of xⁿ=e when n is prime. since the only divisors of n are 1 and itself, we'll have (1 + (n-1)!) solutions.
so, x⁵=e we'll have 25 solutions in S₅.

Right.

we can also predict the number of solutions of x^k=e in S_n when n is prime. the answer will be (1 + (n,k)*(k-1)!).
the number of solutions of x⁴=e in S₅ is 1 + 5*24 = 121.

These are only the 4-cycles.
What about solutions to x²=e in S₅?

the number of solutions of x²=e in S₃ is 1 + 3*1 = 4. those solutions namely are: {e,(1 2),(1 3),(2 3)}.

Yes.