Print the "even" permutations of symmetric group Sn in cyclic notation THE TASK

DEFINITIONS

Consider the points {1,2,3,4,5} and all their permutations. We can find the total number of possible permutations of these 5 points by a simple trick: Imaging filling 5 slots with these points, the first slot will have 5 possible numbers, the second 4 (as one has been used to fill the first slot) the third 3 and so on. Thus the total number of Permutations is 5*4*3*2*1; this would be 5! permutations or 120 permutations. We can think of this as the symmetric group S5, and then Symmetric Group Sn would have n! or (n*n-1*n-2...*1) permutations.

An "even" permutation is one where there is an even number of even length cycles. It is easiest to understand when written in cyclic notation, for example (1 2 3)(4 5) permutes 1-&gt;2-&gt;3-&gt;1 and 4-&gt;5-&gt;4 and has one 3 length cycle (1 2 3) and one 2 length cycle (4 5). When classifying a permutation as odd or even we ignore odd length cycles and say that this permutation [(1 2 3)(4 5)] is odd as it has an odd number {1} of even length cycles.
Even examples:


(1)(2 3)(4 5) = two 2 length cycle | EVEN |
(1 2 3 4 5) = no even length cycles | EVEN | * note that if no even length cycles are present then the permutation is even.


Odd Examples:


(1 2)(3 4 5) = one 2 length cycle | ODD |
(1)(2 3 4 5) = one 4 length cycle | ODD |


As exactly half of the permutations in any Symmetric Group are even we can call the even group the Alternating Group N, So as S5 = 120 A5 = 60 permutations.

NOTATION

Permutations should, for this at least, be written in cyclic notation where each cycle is in different parenthesis and each cycle goes in ascending order. For example (1 2 3 4 5) not (3 4 5 1 2). And for cycles with a single number, such as: (1)(2 3 4)(5) the single / fixed points can be excluded meaning (1)(2 3 4)(5) = (2 3 4). But the identity {the point where all points are fixed (1)(2)(3)(4)(5)} should be written as () just to represent it.

THE CHALLENGE

I would like you to, in as little code possible, take any positive integer as an input {1,2,3,4...} and display all the permutations of the Alternating Group An where n is the input / all the even permutations of Sn. For example:

Input = 3
()
(1 2 3)
(1 3 2)


and 

Input = 4
()
(1 2)(3 4)
(1 3)(2 4)
(1 4)(2 3)
(1 2 3)
(1 3 2)
(1 2 4)
(1 4 2)
(1 3 4)
(1 4 3)
(2 3 4)
(2 4 3)


And as with in the examples I would like for all cycles of one length to be elided, and as for the identity: outputs of  nothing, () {not only brackets but with whatever you are using to show different permutations} or id are acceptable.

EXTRA READING

You can find more information here:


https://en.wikipedia.org/wiki/Permutation
https://en.wikipedia.org/wiki/Alternating_group


GOOD LUCK

And as this is codegolf whoever can print the Alternating Group An's permutations in the shortest bytes wins.
Mathematica, 84 49 31 bytes

GroupElements@*AlternatingGroup


Composition of two functions. Outputs in the form {Cycles[{}], Cycles[{{a, b}}], Cycles[{{c, d}, {e, f}}], ...} representing (), (a b), (c d)(e f), ....