Find all elements x in S4 such that x^4 = e.
(e is the identity)
Cycle notation is advised.
**The attempt at a solution**
I know there are 4 elements in this (1 2 3 4). And I feel like I need to find the order of S which can be found by finding the number of cycles.
Suppose S is a finite group with x elements, then x^n = e for all x in S. That's my start. not sure where to go from there.
isn't it true that any transposition x = (a b) also satisfies x^4 = e?
by the way, if you want to say "all elements of order 4" that is NOT the same thing as "all elements for which x^4 = e".
i'm not saying that is what you mean, i'm just saying that any element of order 2 and the identity also satisfy x^4 = e.
if indeed you mean all elements for which x^4 = e, what about pairs of disjoint 2-cycles? don't just settle for the obvious answers,
think about the problem.
Well would the order here would divide 4. Since there are 4 elements?
x^4 = e means that the order of x is a factor of 4.
All of the elements with order 1, 2, or 4 need to be found.
When an element is written as a product of disjoint cycles, the order of the element is the least common multiple of the lengths of the disjoint cycles.
Elements of order 1 would be:
the identity (1 2 3 4)
Elements of order 2 would be:
(1 2)(3 4)
(1 3)(2 4)
(1 4)(2 3)
(1 2)(3)(4)
(1 3)(2)(4)
(1 4)(2)(3)
(2 3)(1)(4)
(2 4)(1)(3)
(3 4)(1)(2)
Elements of order 4 would be:
(1 2 3 4)
(1 2 4 3)
(1 3 2 4)
(1 3 4 2)
(1 4 2 3)
(1 4 3 2)
So does this mean there are 16 elements so that x^4 = e?