What will be the possible permutations of set S={1,2,3,4}. And how many are of those odd and even?

Question

I have solved it as per my knowledge and understanding. Since there are 4 elements so no. of permutations will be $$4! = 24$$ $$(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (2,1,3,4), (2,1,4,3), (2,3,1,4), (2,3,4,1), (2,4,1,3), (2,4,3,1), (3,2,1,4), (3,2,4,1), (3,1,2,4), (3,1,4,2), (3,4,2,1), (3,4,1,2), (4,2,3,1), (4,2,1,3), (4,3,2,1), (4,3,1,2), (4,1,2,3), (4,1,3,2)$$

Is it correct?

• And for even and odd-

Even: product of even no of transpositions. e.g., $(1,2,3) = (1,2)(1,3)$ is even Odd: product of odd no of transpositions. e.g., $(1,2,3,4) = (1,2)(1,3)(1,4)$ is odd.

• But if I apply this theory in this question then all permutations will be odd. I am confused here, is my solution incorrect? May be there will be more permutations with less elements like $(1,2,3) , (1,2,4)$?

• Also there is a swapping logic! If the numbers are swapped odd times then it is odd and even otherwise

So in this case $(1,2,3,4)$ is even (no swaps) $(3,2,1,4)$ = Swap 1<->3 is odd Also $(4,2,1,3)$ = Swap 3<->4 then swap 1<->4 = even permutation

What will be the answer to my question?

Answer 1

There are twenty four elements in $\mathcal{S}_4$, namely $$\{\operatorname{id}, (1\ 2), (1\ 3), (1\ 4), (2\ 3),\ (2\ 4), (3\ 4), (1\ 2\ 3),\ (1\ 2\ 4), (1\ 3\ 4), (2\ 3\ 4),\ (1\ 3\ 2),\ (1\ 4\ 2),\ (1\ 4\ 3), (2\ 4\ 3),\ (1\ 2\ 3\ 4), (1\ 3\ 4\ 2), (1\ 4\ 2\ 3), (4\ 3\ 2\ 1), (2\ 4\ 3\ 1), (3\ 2\ 4\ 1), (1\ 2)(3\ 4), (1\ 3)(2 \ 4),\ (1\ 4)(2\ 3)\}.$$ In general; there are $n!$ elements in $\mathcal{S}_n.$

Now, list the cycle types of $\mathcal{S}_4$ $$\{(1, 1, 1, 1), (2,1,1),(2,2),(3,1),(4)\}.$$ List how many cycles each cycle type takes, let's call this quantity $\gamma(a)$ where $a$ corresponds to each cycle type. Then $$\gamma(1,1,1,1)=4,\quad\gamma(2,1,1)=3,\quad\gamma(2,2)=2,\quad\gamma(3,1)=2,\quad\gamma(4)=1.$$ Then, a permutation is even if $n-\gamma(a)$ is even. So, the even permutations are the ones of cycle type $(1,1,1,1)$, $(2,2)$, $(3,1)$. This gives us a total of twelve even permutations, namely $$\{\operatorname{id}, (1\ 2)(3\ 4), (1\ 3)(2 \ 4),\ (1\ 4)(2\ 3),(1\ 2\ 3),\ (1\ 2\ 4), (1\ 3\ 4), (2\ 3\ 4),\ (1\ 3\ 2),\ (1\ 4\ 2),\ (1\ 4\ 3), (2\ 4\ 3)\},$$ with the other twelve permutations being odd.

In general, there are $n!/2$ even and odd permutations in $\mathcal{S}_n$.

Answer 2

Don't be put off! You have made a very simple mistake which might be easily rectified.

I haven't checked your list for completeness, but I suspect it does include all the permutations, you've just mixed up two different notations.

I assume for example that in the list $(4,2,3,1)$ is the permutation mapping $1\mapsto 4$, $2\mapsto 2$, $3\mapsto 3$, $4\mapsto 1$? The usual (or cycle) notation for this would be $(1,4)$, an odd permutation.

As you have written $(2,3,1,4)$, would usually be written $(1,2,3)=(1,2)(1,3)$ (or $(1,3)(1,2)$ depending on the convention you use), an even permutation.

See this wiki page for a brief description of the cycle notation.

As for how many are even and how many are odd, you must have proven that a cycle is either even or odd, not both. So if $x$ is an odd permutation, then it is the product of an odd number of transpositions, so $(1,2)x$ is a product of an even number of transpositions so is even. This map $x\mapsto (1,2)x$ is a bijection between the set of odd and the set of even permutations. What does this tell you about how much there must be of each?