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Let $\pi$ be the partition of $n=a_1+a_2+...+a_r$, where $a_1\geq a_2 \geq ....\geq a_r\gt0$ and $\pi$’ be the partition, $n=b_1+b_2+....+b_s”.$Prove that $\pi$’ is the conjugate of $\pi$ if and only if $r=b_1$ and $s=a_1$ and $b_j$ is the number of parts in the partition of $\pi$ that are $\geq j$ for $j=1,2,....s$

What i did so far:

If we consider the ferrers diagram of the partition $n=a_1+a_2+...+a_r$ then $a_r$ denotes the number of rows of the representation and r denotes the number of columns in the diagram. Similarly for the other representation $b_s$ denotes the number of rows of the representation while s denotes the number of columns of the representation.

Now if the conjugate of $n=b_1+b_2+...b_s$ were to be $n=a_1+a_2+...+a_r$ then the largest of all $b_i$’s i.e $b_1$ will be the number of columns, i.e $b_1=r$. Similarly the number of terms i.e s will be the number of rows, i.e $s=a_1$.

So, can anyone help me in proceeding towards the converse part ?, also is $s=a_1$ and $r=b_1$ enough to state that $\pi$ is the conjugate of $\pi$?

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