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Conjecture not true.
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metallicmural99
metallicmural99
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Context: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by $$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size. (see this post)

Question: Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.

EDIT:

The conjecture as it stands is not true, as per one of the answers posted below. To have roughly 1/2 of the parts as 1's, the number of parts also needs to be roughly a 1/2 of $n$ or so, it seems.

Context: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by $$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size. (see this post)

Question: Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.

Context: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by $$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size. (see this post)

Question: Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.

EDIT:

The conjecture as it stands is not true, as per one of the answers posted below. To have roughly 1/2 of the parts as 1's, the number of parts also needs to be roughly a 1/2 of $n$ or so, it seems.

Source Link
metallicmural99
metallicmural99
  • 211
  • 1
  • 7

Maximum conjugacy class size in $S_n$ with fixed number of cycles

Context: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by $$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size. (see this post)

Question: Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.