Decomposition of products of monomial symmetric polynomials into sums of them I'm trying to make sense of the answer given in: this question
I am stuck at the phrase 'where the partitions $\gamma$ result from adding, respectively, from $\alpha$ all distinct partitions obtained by permuting in all possible ways the parts of $\beta$.'
If I permute (in all possible ways) the parts of a partition, then I obtain a set of integers that is not weakly decreasing anymore; adding that part-by-part to a proper partition does again not conserve the property 'weakly decreasing'. 
Q: how should I interpret this?
Yet, there is a tempting structure in this product decomposition that I would like to understand. 
Example: $m[1,1]^2= m[2,2]+2 m[2,1,1]+6 m[1,1,1,1]$ (* using 4 variables *)
added 19/05/2013 15:42 CET: @MarcVanLeeuwen: thanks Marc, I got it now. The 'Dim' function was new to me, it turns out to be just the multinomial of the contents (of the zero-padded partitions). 
 A: It is natural to try to manipulate symmetric polynomials as one does polynomials, representing them as a sum of monomials, and to take advantage of the symmetry by representing a whole orbit of monomials only once, as a minimal symmetric polynomial$~m_\lambda$. Also one can now leave the number of indeterminates unspecified, and thus effectively manipulate symmetric functions. This being said, I should warn that the algorithms resulting form this head-on approach to symmetric functions are neither particularly beautiful nor efficient; representation on other bases with more pleasant theoretical properties seems preferable.
Anyway, the best way to understand the multiplication of $m_\mu$ and $m_\nu$ is to try deriving the rule for the coefficients yourself, since only this gives you the right feeling for what kind of complications are needed; reading a ready-made description of the multiplication risks misinterpreting the subtle notions involved. Here I will try to deduce the rule. The first thing to realise it that if $l(\lambda)$ denotes the number of nonzero parts of$~\lambda$, then in the product of symmetric functions $m_\mu m_\nu$ one get terms $m_\lambda$ with $l(\lambda)$ going up to $l(\mu)+l(\nu)$, so one should consider at least that number of indeterminates to faithfully represent that product as a polynomial. If you are only interested in multipliying symmetric polynomials in a fixed number of indeterminates one can of course use that number, which amounts to ignoring the $m_\lambda$ where $l(\lambda)$ exceeds that number.
If $n$ is the number of indeterminates used, it is useful to consider scalar multiples of the $m_\lambda$ obtained by a sum over the whole group $S_n$, namely $$M_\lambda(n)=c(\lambda,n)m_\lambda=\sum_{\pi\in S_n}X^{\pi(\lambda)},$$ where $\def\N{\mathbf N}\pi(\lambda)\in\N^n$ is the result of permuting the parts of$~\lambda$ (including $n-l(\lambda)$ parts zero) by $\pi$; the number $c(\lambda,n)$ is the size of the stabiliser of $\lambda$ in this action, the product of the factorials of the multiplicities of the distinct parts of $\lambda$, including $(n-l(\lambda))!$ for the parts equal to $0$ involved. Now clearly
$$
  M_\mu(n)M_\nu(n)=\sum_{\pi,\sigma\in S_n}X^{\pi(\mu)+\sigma(\nu)}
 =\sum_\lambda C^\lambda_{\mu,\nu}M_\lambda(n),
$$
and once these coefficients $C^\lambda_{\mu,\nu}$ are determined one deduces that 
$$
  m_\mu m_\nu=\sum_\lambda c^\lambda_{\mu,\nu}m_\lambda,\qquad
 \text{where }
  c^\lambda_{\mu,\nu}=\frac{c(\lambda,n)}{c(\mu,n)c(\nu,n)}C^\lambda_{\mu,\nu}.
$$
The defining equation for the coefficients $C^\lambda_{\mu,\nu}$ is
$$
  \sum_\lambda C^\lambda_{\mu,\nu}\sum_{\pi\in S_n}X^{\pi(\lambda)}
 =\sum_{\pi,\sigma\in S_n}X^{\pi(\mu)+\sigma(\nu)}
 =\sum_{\pi,\sigma\in S_n}X^{\pi(\mu+\pi^{-1}(\sigma(\nu)))},
$$
which by symmetry can be simplified to
$$
  \sum_\lambda C^\lambda_{\mu,\nu}X^\lambda
 =\sum_{\rho\in S_n}X^{(\mu+\rho(\nu))^+},
$$
where for a composition $\alpha$ the notation $\alpha^+$ means that partition obtained by sorting its parts decreasingly.
Now we can simplify this further by factoring the stabiliser subgroup of $\nu$, to
$$
  \sum_\lambda \frac{C^\lambda_{\mu,\nu}}{c(\nu,n)}X^\lambda
 =\sum_{\alpha\in S_n(\nu)}X^{(\mu+\alpha)^+},
$$
where $S_n(\nu)$ is the permutation orbit, without repetitions. In the end, the recipe is
$$
  c^\lambda_{\mu,\nu}=\frac{c(\lambda,n)}{c(\mu,n)}c'(\lambda,\mu,\nu)
\quad\text{where}\quad
  \sum_{\alpha\in S_n(\nu)}X^{(\mu+\alpha)^+}=\sum_\lambda c'(\lambda,\mu,\nu)X^\lambda;
$$
the right hand side asks for adding all distinct permutations of $\nu$ (which was padded with zero) to $\mu$, sorting the results decreasingly and grouping by equal outcome$~\lambda$.
