0

$S_4$ is the set of all permutations of length 4

Let $a=\binom{1\,2\,3\,4}{3\,2\,1\,4}$

I found that $a$ is an odd permutation and I want to demonstrate that the product of the permutations is even but Im not sure if this is true and I have no idea how to demonstrate that.

oren revenge
  • 566
  • 4
  • 13
  • 2
    Product of what permutations? – almagest May 29 '16 at 08:35
  • If you want to demonstrate the product of two odd permutations is even the use the definition. like in your case $a=(1,3)$ which has odd number of 2 cycles so is an odd permutation. In general any permutation $\tau$ can be split up into $2$ cycles, and if the number of $2$ cycles is odd, we say $\tau$ is odd, if number of $2$ cycles is even, then $\tau$ is called an even permutation – C.S. May 29 '16 at 08:44
  • @S.C.: How about $({}^1_2,{}^2_3,{}^3_4,{}^4_1)$? It consists of a single 4-cycle and no 2-cycles, so according to your definition it should be even. But actually it's odd. What you want to say is that it is the product of an odd number of transpositions. It is true that a transposition and a 2-cycle is the same, but writing a permutation as a product of cycles is not the same as writing it as a product of transpositions. "Split into cycles" usually implies that the cycles in question are disjoint. – hmakholm left over Monica May 29 '16 at 08:53
  • @HenningMakholm $(1 \ 2 \ 3 \ 4) = (1\ 2) (1 \ 3 ) (1 \ 4 )$ which says its odd, since it has been split up into odd number of 2 cycles. and hence odd. Where is the error in my definition – C.S. May 29 '16 at 09:21
  • @S.C.: The error is that what you show is not the cycle decomposition of the permutation, and you should not speak about expressing a permutation as a product of transpositions with the language for splitting into cycles. That just causes unnecessary confusion. – hmakholm left over Monica May 29 '16 at 09:22
  • @HenningMakholm So how will u decompose $(1\ 2 \ 3 \ 4 )$ as disjoint cycles? you can't right. It just is a single cycle isn't it? – C.S. May 29 '16 at 09:26
  • @S.C.: It's just a single cycle, yes. But that doesn't mean it doesn't have a cycle decomposition. The cycle decomposition of $(1,2,3,4)$ is $(1,2,3,4)$ itself. – hmakholm left over Monica May 29 '16 at 09:31
  • @HenningMakholm Yeah thats what i meant :). Thanks for taking time to clarify :) – C.S. May 29 '16 at 09:32

1 Answers1

1

I think what you're saying is that if you take all the permutations in $S_4$ and multiply them together in whichever order, the result is never equal to your $a$.

Your basic plan is sound -- $a$ is indeed an odd permutation, and it will turn out that a product of all the permutations is even.

The product of all the permutations is even if and only if the total number of odd permutations in the product is even.

There are various ways of showing that. A quick and slick one is would be to know that exactly half of the elements of $S_n$ are odd (for $n\ge 2$), compute the size of $S_4$ and see that half of that number is even.

(Proof that exactly half of all permutations are odd: Fix some transposition -- say, your $a$ -- and then observe that if $\sigma$ is an even permutation, then $a\sigma$ is odd, and if $\tau$ is an even permutation, then $a\tau$ is an odd permutation such that $a(a\tau)=\tau$. So left multiplication by $a$ is a bijection between the even and the odd permutations, so there are equally many of them).

hmakholm left over Monica
  • 286,031