LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)audited 2026-08-28

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Conjugating a partition twice returns the original partition

Statement

For every partition $λ$ , one has

$(λ^{'})^{'} = λ .$

Facts & Assumptions

Given: a partition $λ$ and its Ferrers diagram.

[F1]

The part $λ_{j}^{'}$ is the number of rows of $λ$ of length at least $j$ (Ferrers and Young diagrams, conjugate partitions, self-conjugacy, and the Durfee square).

Proof

technique · direct

1.1

A cell $(i, j)$ belongs to the Ferrers diagram of $λ$ exactly when the $i$ -th row has length at least $j$ , and by [F1] this is equivalent to the $j$ -th row of $λ^{'}$ having length at least $i$ . Therefore the cells of $λ^{'}$ are exactly the transpose $(j, i)$ of the cells of $λ$ .

2.1

Transposing the same finite set of cells a second time sends $(j, i)$ back to $(i, j)$ , so the twice-conjugated diagram is the original diagram. Hence its row lengths are again $λ$ , that is, $(λ^{'})^{'} = λ$ .

step 1.1∎

Depends on

Ferrers and Young diagrams, conjugate partitions, self-conjugacy, and the Durfee square

Used by

Conjugating (4,2,1) does not produce an odd-part partition Counterexample
Conjugation pairs the partitions of 6 by swapping length and largest part Example
FALSE: conjugation itself is the distinct-parts to odd-parts bijection False statement
Partitions with k parts are equinumerous with partitions whose largest part is k Theorem

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Stephen Melczer, An Invitation to Enumeration, Chapter 9: Integer Partitions (standard reference, not scraped)