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While experimenting with symmetric functions, I noticed the following equality of subrings of the ring of symmetric functions: $$\mathbb{Z}[(n-1)!\cdot p_n \ |\ n \ge 1] = \mathbb{Z}[n!\cdot h_n \ |\ n \ge 1],$$ where the $p_n$ are the power-sum symmetric functions and the $h_n$ are the complete homogeneous symmetric functions. Is this fact known, or can anyone think of a simple proof of it?

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    $\begingroup$ One of the famous identities in symmetric function theory can be rewritten as $\sum\limits_{m \geq 1} \left(m-1\right)! p_m \cdot \dfrac{t^m}{m!} = \log \sum\limits_{m \geq 0} m! h_m \cdot \dfrac{t^m}{m!}$. Of course, the $\dfrac{t^m}{m!}$'s here can be viewed as the basis elements of the divided powers algebra. This makes me suspect that $\log$ on the divided powers algebra (at least over $\mathbb{Z}$) is a bijection (under appropriate constant term conditions), and the relation between a divided-powers series $f$ and its logarithm $\log f$ is such that the $\mathbb{Z}$-subalgebra ... $\endgroup$ Feb 18, 2022 at 0:47
  • $\begingroup$ ... generated by the coefficients of the former is the $\mathbb{Z}$-subalgebra generated by the coefficients of the latter (this should actually follow from the "bijection" part, because you can assume that either of these two subalgebras is your base ring). This ought to be known, right? $\endgroup$ Feb 18, 2022 at 0:47
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    $\begingroup$ Alternatively, you can multiply the recursive Newton identity $kh_k = \sum\limits_{i=1}^k h_{k-i} p_i$ by $\left(k-1\right)!$ to obtain $k! h_k = \sum\limits_{i=1}^k \dbinom{k-1}{i-1} \left(k-i\right)! h_{k-i} \left(i-1\right)! p_i$, and use this to prove your claim by the standard induction argument that you would normally use. Nice exercise! $\endgroup$ Feb 18, 2022 at 0:57

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