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I would like to know whether the notion of automorphism of the set of partitions of a positive integer $n$ has been considered so far or not. To make things clearer, I say that a partition of $n$ in $k$ summands $s_1,s_2,...s_k$ sorted in decreasing order has signature $(a_1,a_2,...,a_m)$ if and only if $\forall i\in\{1,\cdots,m\},\ \ a_{i}>0$ and $\displaystyle{\sum_{i=1}^{m}a_{i}=k}$ and each $a_i$ is the multiplicity of a summand: $a_1$ is the number of times the largest summand appears, $a_2$ the number of times the second largest summand appears, and so on.

I thus call "automorphism of the set of partitions of $n$" any bijective map sending any partition of $n$ in $k$ summands of signature $(a_1,a_2,...a_m)$ to a partition of $n$ in $k$ summands of signature $(a_1, a_2,...,a_m)$. For example, one can consider a map sending the partition $(6,5,1)$ of $12$ to the partition $(6,4,2)$.

If I'm not mistaken, the group of automorphisms of the set of partitions of $n$ is a subgroup $Aut_{Part}(n)$ of $S_{p(n)}$, where $p(n)$ is the number of partitions of $n$. What can be said about this group? Is it necessarily abelian? Solvable? Any interesting reference available on-line freely about related subjects?

Many thanks in advance.

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    $\begingroup$ Your definition of the signature of a partition does not mention the partition except for the number of parts $k$. Something is wrong. $\endgroup$
    – Douglas Zare
    Commented Jan 24, 2016 at 18:30
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    $\begingroup$ don't you mean that your group is a subgroup of $S_{p(n)}$, where $p(n)$ is the number of partitions of $n$? $\endgroup$
    – Marcel
    Commented Jan 24, 2016 at 18:37
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    $\begingroup$ I still don't know what the signature is of a partition. How do you get $(1,1)$ from the partition $6+6$ of $12$? $\endgroup$
    – Douglas Zare
    Commented Jan 24, 2016 at 18:41
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    $\begingroup$ Please clarify: "any bijective map": a map between which set and which set? A bijection from a set to itself is called a permutation of the set, and you should specify the set. $\endgroup$
    – YCor
    Commented Jan 24, 2016 at 18:43
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    $\begingroup$ Please actually define what you mean. $\endgroup$
    – Douglas Zare
    Commented Jan 24, 2016 at 18:44

1 Answer 1

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Let $P(n,\alpha)$ be the set of partitions of $n$ with signature $\alpha$. Such a set is invariant under your group, and any permutation of its elements is an automorphism. Your group is just $$G=\Pi_\alpha S_{|P(n,\alpha)|}.$$

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  • $\begingroup$ You probably mean $\prod_\alpha$? Or I ignore some use of the $\otimes$ sign. Anyway I have the same understanding of the question (which makes it trivial) $\endgroup$
    – YCor
    Commented Jan 24, 2016 at 18:45
  • $\begingroup$ Thank you Marcel for both understanding and clarifying what I meant. But I must admit I'm in the same situation as YCor. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 24, 2016 at 18:48
  • $\begingroup$ I mean direct product of groups. How do you denote this? $\endgroup$
    – Marcel
    Commented Jan 24, 2016 at 18:49
  • $\begingroup$ $\Pi_\alpha$. I agree with this answer even though I don't know what type of object $\alpha$ is, a composition or a partition. $\endgroup$
    – Douglas Zare
    Commented Jan 24, 2016 at 18:50
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    $\begingroup$ Besides being what I said, it is a partition of $k$. So $(3,2,2)$ and $(5,1,1)$ are partitions of $7$, both with signature $(2,1)$. $\endgroup$
    – Marcel
    Commented Jan 24, 2016 at 18:58

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