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Assume that $\sigma \in S_n$ and $\sigma = \alpha_1 \dots\alpha_s$ is the decomposition of $\sigma$ into disjoint cycles, such that all of the members of $\{1,2,\dots,n\}$ are appeared in the members of $\alpha_1, \dots, \alpha_s$.

Is this statement true?

The minimum number of transpositions needed to decompose $\sigma$ is $n-S$.

Paolo Molina
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3 Answers3

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amakelov explains why $n-S$ transpositions are sufficient. To show that $n-S$ transpositions are necessary, one proves that for a permutation $\sigma$ and a transposition $\tau$, $\sigma\tau$ has one more, or one fewer cycle than $\sigma$. As the identity permutation has $n$ cycles, a product of $k$ transpositions has at least $n-k$ cycles. So if $k<n-S$, a product of $k$ transpositions has more than $S$ cycles.

Angina Seng
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It's true because every cycle of length $k$ can be decomposed as the product of $k-1$ transpositions. The way to see that is to observe that $$(1,2,\ldots,k)= (1,2)(2,3)\ldots(k-1,k)$$ which can be checked directly from the formula (each $i$ goes to $i+1$ modulo $k$)

amakelov
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$d=n-s$ is called the decrement of the permutation $\sigma$. It not always is equal to the minimal number of transpositions needed to write $\sigma$.

Example. Assume for simplicity $n=2$, i.e., we are in $S_2$. The identity permutation $e=(1)(2)$ is of degree $n=2$, and it has $2$ disjoint cycles. Hence its decrement is $d=2-2=0$. But $e$ cannot be written by zero transpositions.

By definition a transposition is a permutation of type $(ij)$, and hence we have to write, say, $e=(12)(12)$. At least two transpositions are needed. You cannot write, say, $e=e$ and call it transpositions decomposition by zero transpositions because $e$ is not a transposition.

By the way, for this very same reason we in the Fundamental theorem of arithmetic require that the integer $n$ must be greater than $1$ to have a decomposition as a product of primes. For, the equality $1=1$ cannot be interpreted as a product of zero primes.

My students had seen this page and discussed the issue with me. Hence I add this remark here.

V. Mikaelian
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  • $e$ is not a transposition, but it can be decomposed onto $0$ transpositions. Denoting a sequence of adjacent transpositions $\tau_1, \tau_2, ..., \tau_n$ as $[\tau_1, \tau_2, ..., \tau_n]$ we can write $e = [\space]$. The result of $[\space]$ is $e$. – Alex C Apr 05 '24 at 04:58
  • Alex C, I see your point: adding some extra formalism to transpositions decomposition one may write $e=[ ]$ to make sure the formula $n-s$ still holds.

    The only purpose of this new formalism is to justify $n-s$, and it makes the general definition of transpositions decomposition somewhat more complicated...

     – V. Mikaelian Apr 06 '24 at 16:14
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    I am saying that the result of application of $0$ transpositions is the identity permutation, meaning the identity permutation can be decomposed to $0$ transpositions. The only purpose of the formalism was to illustrate that simple statement. – Alex C Apr 07 '24 at 03:46
  • The answer turned out to be long, and I separate it to 3 parts.

    Part 1 of 3:

    Dear Alex, we both understand that what we discuss is more about formalism, rather than about actual mathematical truth. Hence, the question is: Is the product of zero transpositions the identity according to popular formalism adopted in Algebra, in general? The answer may not be positive necessarily. For example:

     – V. Mikaelian Jul 17 '24 at 06:11
  • Part 2 of 3:

    1) How we define the identity element in a free group $F_X$ on the alphabet $X$? We first define the words of length at least one on $X$, and then we separately define the identity element $1 \in F_X$ as a new empty word (to which we assign length $0$). Nobody defines the identity element in $F_X$ as a product of zero letters from the alphabet.

     – V. Mikaelian Jul 17 '24 at 06:11
  • Part 3 of 3:

    2) How we set the representation of invertible matrices as products of (three types of) elementary matrices? Nobody tells that the identity matrix $I$ is a product of zero elementary matrices. If that were the case, then one would have to admit that there is a matrix that can be written as a product of zero other matrices...

     – V. Mikaelian Jul 17 '24 at 06:12
  • I am afraid there is a confusion between group action and group operation in your examples. Please rephrase them in terms of transformations of underlying sets to match the subject of the decomposition law we discuss. – Alex C Jul 18 '24 at 05:42