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\begin{document}
\section*{Math 701 Fall 2026, Lecture 3: The classical families $e,h,p$}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}

We continue where we left off in Lecture 2.

\setcounter{section}{1}\setcounter{subsection}{2}\setcounter{theo}{3}

\begin{proof}
[Proof of the identity (\ref{eq.thm.symp.newton.eh}) from Lecture 2.]Parts
\textbf{(a)} and \textbf{(c)} of Proposition \ref{prop.symp.eh-genfun} yield%
\begin{align*}
\prod_{i=1}^{N}\left(  1-tx_{i}\right)   &  =\sum_{n\in\mathbb{N}}\left(
-1\right)  ^{n}t^{n}e_{n}\ \ \ \ \ \ \ \ \ \ \text{and}\\
\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}  &  =\sum_{n\in\mathbb{N}}t^{n}h_{n}%
\end{align*}
in the power series ring $\mathcal{P}\left[  \left[  t\right]  \right]  $.
Multiplying these two equalities, we find\footnote{All our manipulations with
power series, such as expanding products of infinite sums or interchanging
summation signs, are justified by the coefficientwise topology on the ring of
formal power series. This is the topology with respect to which a sequence
$\left(  f_{0},f_{1},f_{2},\ldots\right)  $ (or, more generally, a net
$\left(  f_{d}\right)  _{d\in D}$) of power series converges to a power series
$f$ if and only if for each $n\in\mathbb{N}$, all sufficiently high
$d\in\mathbb{N}$ (or, in the case of a net, all sufficiently high $d\in D$)
satisfy $\left[  t^{n}\right]  f_{d}=\left[  t^{n}\right]  f$ (where $\left[
t^{n}\right]  g$ denotes the $t^{n}$-coefficient of a power series $g$). The
\textquotedblleft sufficiently high\textquotedblright\ here can depend on $n$.
All our sums converge with respect to this topology. For a more elementary
explanation of this, see \cite[Chapter 3]{21s} (for the univariate case).}%
\begin{align*}
1  &  =\left(  \sum_{n\in\mathbb{N}}\left(  -1\right)  ^{n}t^{n}e_{n}\right)
\left(  \sum_{n\in\mathbb{N}}t^{n}h_{n}\right) \\
&  =\left(  \sum_{i\in\mathbb{N}}\left(  -1\right)  ^{i}t^{i}e_{i}\right)
\left(  \sum_{j\in\mathbb{N}}t^{j}h_{j}\right) \\
&  =\underbrace{\sum_{i\in\mathbb{N}}\ \ \sum_{j\in\mathbb{N}}}_{=\sum
_{n\in\mathbb{N}}\ \ \sum_{\substack{\left(  i,j\right)  \in\mathbb{N}%
^{2};\\i+j=n}}}\left(  -1\right)  ^{i}\underbrace{t^{i}e_{i}t^{j}h_{j}%
}_{=e_{i}h_{j}t^{i+j}}\\
&  =\sum_{n\in\mathbb{N}}\ \ \sum_{\substack{\left(  i,j\right)  \in
\mathbb{N}^{2};\\i+j=n}}\left(  -1\right)  ^{i}e_{i}h_{j}\underbrace{t^{i+j}%
}_{\substack{=t^{n}\\\text{(since }i+j=n\text{)}}}\\
&  =\sum_{n\in\mathbb{N}}\left(  \sum_{\substack{\left(  i,j\right)
\in\mathbb{N}^{2};\\i+j=n}}\left(  -1\right)  ^{i}e_{i}h_{j}\right)  t^{n}.
\end{align*}
Comparing $t^{n}$-coefficients, we see that each $n\in\mathbb{N}$ satisfies%
\[%
\begin{cases}
1, & \text{if }n=0;\\
0, & \text{if }n>0
\end{cases}
\ \ =\sum_{\substack{\left(  i,j\right)  \in\mathbb{N}^{2};\\i+j=n}}\left(
-1\right)  ^{i}e_{i}h_{j}=\sum_{i=0}^{n}\left(  -1\right)  ^{i}e_{i}h_{n-i}%
\]
(since the pairs $\left(  i,j\right)  \in\mathbb{N}^{2}$ satisfying $i+j=n$
are simply the pairs $\left(  i,n-i\right)  $ for $i\in\left\{  0,1,\ldots
,n\right\}  $). If $n$ is a positive integer, then the left-hand side of this
equality is simply $0$, so the whole equality rewrites as%
\[
0=\sum_{i=0}^{n}\left(  -1\right)  ^{i}e_{i}h_{n-i}=\sum_{j=0}^{n}\left(
-1\right)  ^{j}e_{j}h_{n-j}.
\]
Thus, (\ref{eq.thm.symp.newton.eh}) is proved.
\end{proof}

The proofs of the other two Newton--Girard identities (that is,
(\ref{eq.thm.symp.newton.ep}) and (\ref{eq.thm.symp.newton.hp})) rely on the
notion of the \textbf{logarithmic derivative} of an FPS\footnote{The
abbreviation \textquotedblleft\textbf{FPS}\textquotedblright\ means
\textquotedblleft formal power series\textquotedblright.}:

\begin{definition}
\label{def.fps.loder}For any FPS $f\in K\left[  \left[  t\right]  \right]  $
with constant term $1$, define the \textbf{logarithmic derivative}
$\operatorname*{loder}f$ of $f$ by%
\[
\operatorname*{loder}f=\dfrac{f^{\prime}}{f}\in K\left[  \left[  t\right]
\right]  .
\]
(As we recall, $f^{\prime}$ denotes the derivative of $f$ with respect to $t$.
Division by $f$ is allowed since $f$ has constant term $1$.)
\end{definition}

\begin{proposition}
[rules for logarithmic derivatives]\label{prop.fps.loder}Logarithmic
derivatives satisfy the following rules:

\begin{enumerate}
\item[\textbf{(a)}] For any two FPSs $f$ and $g$ with constant terms $1$, we
have%
\[
\operatorname*{loder}\left(  fg\right)  =\operatorname*{loder}%
f+\operatorname*{loder}g.
\]


\item[\textbf{(b)}] For any $k$ FPSs $f_{1},f_{2},\ldots,f_{k}$ with constant
terms $1$, we have%
\[
\operatorname*{loder}\left(  \prod_{i=1}^{k}f_{i}\right)  =\sum_{i=1}%
^{k}\operatorname*{loder}f_{i}.
\]


\item[\textbf{(c)}] For any FPS $f$ with constant term $1$, we have%
\[
\operatorname*{loder}\dfrac{1}{f}=-\operatorname*{loder}f.
\]


\item[\textbf{(d)}] If $K$ is a commutative $\mathbb{Q}$-algebra, then any FPS
$f$ with constant term $1$ satisfies%
\[
\operatorname*{loder}f=\left(  \log f\right)  ^{\prime}.
\]
Here, $\log f$ denotes the composition of the FPSs $\log\left(  1+t\right)
=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}t^{n}$ and $f-1$ (for
details, see \cite[Definition 3.7.6]{21s}, where $\log$ is denoted by
$\operatorname*{Log}$).
\end{enumerate}
\end{proposition}

\begin{proof}
\textbf{(a)} By the definition of a logarithmic derivative, we have%
\begin{align*}
\operatorname*{loder}\left(  fg\right)   &  =\dfrac{\left(  fg\right)
^{\prime}}{fg}=\dfrac{f^{\prime}g+fg^{\prime}}{fg}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the product rule}\right) \\
&  =\underbrace{\dfrac{f^{\prime}}{f}}_{=\operatorname*{loder}f}%
+\underbrace{\dfrac{g^{\prime}}{g}}_{=\operatorname*{loder}g}%
=\operatorname*{loder}f+\operatorname*{loder}g.
\end{align*}
This proves part \textbf{(a)}. \medskip

\textbf{(b)} Induct on $k$. The \textit{base case} ($k=0$) is just saying
$\operatorname*{loder}1=0$, which is clear (since $1^{\prime}=0$). The
\textit{induction step} uses part \textbf{(a)}. \medskip

\textbf{(c)} Part \textbf{(a)} shows that $\operatorname*{loder}$ is a group
morphism from the multiplicative group of FPSs with constant term $1$ to the
additive group of all FPSs. But group morphisms respect inverses. So
$\operatorname*{loder}\dfrac{1}{f}=-\operatorname*{loder}f$. \medskip

\textbf{(d)} Quick and dirty argument: the chain rule yields%
\[
\left(  \log f\right)  ^{\prime}=\underbrace{\log^{\prime}f}_{=\dfrac{1}{f}%
}\cdot\,f^{\prime}=\dfrac{1}{f}\cdot f^{\prime}=\dfrac{f^{\prime}}%
{f}=\operatorname*{loder}f.
\]
\textquotedblleft But wait\textquotedblright, you may say, \textquotedblleft
is this allowed when $\log$ is not an actual FPS?\textquotedblright. And
indeed, this calculation is not quite kosher for this exact reason; instead,
we have to decompose $\log f$ as the composition of the actual FPSs
$\log\left(  1+t\right)  $ and $f-1$ and then apply the chain rule to these
two FPSs. You are invited to make this change yourself, or look up the proof
in \cite[Proposition 3.7.13]{21s}.
\end{proof}

\begin{proof}
[Proof of (\ref{eq.thm.symp.newton.hp}) from Lecture 2.]We shall apply
Proposition \ref{prop.fps.loder} with $\mathcal{P}$ in place of $K$.

Proposition \ref{prop.symp.eh-genfun} \textbf{(c)} in Lecture 2 says that%
\[
\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}=\sum_{n\in\mathbb{N}}t^{n}h_{n}.
\]
Applying the logarithmic derivative to this equality, we obtain%
\[
\operatorname*{loder}\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}=\operatorname*{loder}%
\left(  \sum_{n\in\mathbb{N}}t^{n}h_{n}\right)  =\dfrac{\left(  \sum
_{n\in\mathbb{N}}t^{n}h_{n}\right)  ^{\prime}}{\sum_{n\in\mathbb{N}}t^{n}%
h_{n}}=\dfrac{\sum_{n>0}nt^{n-1}h_{n}}{\sum_{n\in\mathbb{N}}t^{n}h_{n}},
\]
since the definition of a derivative yields $\left(  \sum_{n\in\mathbb{N}%
}t^{n}h_{n}\right)  ^{\prime}=\sum_{n>0}nt^{n-1}h_{n}$. Thus,%
\begin{align*}
\dfrac{\sum_{n>0}nt^{n-1}h_{n}}{\sum_{n\in\mathbb{N}}t^{n}h_{n}}  &
=\operatorname*{loder}\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}\\
&  =\sum_{i=1}^{N}\underbrace{\operatorname*{loder}\dfrac{1}{1-tx_{i}}%
}_{\substack{=\operatorname*{loder}\left(  \left(  1-tx_{i}\right)
^{-1}\right)  \\=-\operatorname*{loder}\left(  1-tx_{i}\right)  \\\text{(by
Proposition \ref{prop.fps.loder} \textbf{(c)})}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition \ref{prop.fps.loder} \textbf{(b)}}\right) \\
&  =-\sum_{i=1}^{N}\underbrace{\operatorname*{loder}\left(  1-tx_{i}\right)
}_{=\dfrac{\left(  1-tx_{i}\right)  ^{\prime}}{1-tx_{i}}=\dfrac{-x_{i}%
}{1-tx_{i}}}=-\sum_{i=1}^{N}\dfrac{-x_{i}}{1-tx_{i}}\\
&  =\sum_{i=1}^{N}\underbrace{\dfrac{x_{i}}{1-tx_{i}}}_{\substack{=x_{i}%
\left(  1+tx_{i}+\left(  tx_{i}\right)  ^{2}+\cdots\right)  \\\text{(by the
geometric series formula)}}}=\sum_{i=1}^{N}\underbrace{x_{i}\left(
1+tx_{i}+\left(  tx_{i}\right)  ^{2}+\cdots\right)  }_{\substack{=x_{i}%
+tx_{i}^{2}+t^{2}x_{i}^{3}+\cdots\\=t^{0}x_{i}^{1}+t^{1}x_{i}^{2}+t^{2}%
x_{i}^{3}+t^{3}x_{i}^{4}+\cdots}}\\
&  =\sum_{i=1}^{N}\left(  t^{0}x_{i}^{1}+t^{1}x_{i}^{2}+t^{2}x_{i}^{3}%
+t^{3}x_{i}^{4}+\cdots\right) \\
&  =t^{0}\underbrace{\sum_{i=1}^{N}x_{i}^{1}}_{=p_{1}}+\,t^{1}\underbrace{\sum
_{i=1}^{N}x_{i}^{2}}_{=p_{2}}+\,t^{2}\underbrace{\sum_{i=1}^{N}x_{i}^{3}%
}_{=p_{3}}+\,t^{3}\underbrace{\sum_{i=1}^{N}x_{i}^{4}}_{=p_{4}}+\cdots\\
&  =t^{0}p_{1}+t^{1}p_{2}+t^{2}p_{3}+t^{3}p_{4}+\cdots=\sum_{n>0}t^{n-1}p_{n}.
\end{align*}
That is,%
\begin{align*}
\sum_{n>0}nt^{n-1}h_{n}  &  =\left(  \sum_{n>0}t^{n-1}p_{n}\right)  \left(
\sum_{n\in\mathbb{N}}t^{n}h_{n}\right)  =\left(  \sum_{j>0}t^{j-1}%
p_{j}\right)  \left(  \sum_{i\in\mathbb{N}}t^{i}h_{i}\right) \\
&  =\sum_{j>0}\ \ \sum_{i\in\mathbb{N}}\underbrace{t^{j-1}p_{j}t^{i}h_{i}%
}_{=t^{i+j-1}p_{j}h_{i}}=\sum_{j>0}\ \ \sum_{i\in\mathbb{N}}t^{i+j-1}%
p_{j}h_{i}=\underbrace{\sum_{j>0}\ \ \sum_{n\geq j}}_{=\sum_{n>0}%
\ \ \sum_{j=1}^{n}}\underbrace{t^{n-j+j-1}}_{=t^{n-1}}p_{j}h_{n-j}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted
}n-j\text{ for }i\text{ in the inner sum}\right) \\
&  =\sum_{n>0}\ \ \sum_{j=1}^{n}t^{n-1}p_{j}h_{n-j}=\sum_{n>0}\left(
\sum_{j=1}^{n}p_{j}h_{n-j}\right)  t^{n-1}.
\end{align*}
Comparing $t^{n-1}$-coefficients, we get%
\[
nh_{n}=\sum_{j=1}^{n}p_{j}h_{n-j}=\sum_{j=1}^{n}h_{n-j}p_{j}.
\]
This is precisely the 3rd Newton--Girard identity (\ref{eq.thm.symp.newton.hp}).
\end{proof}

The proof of the 2nd Newton--Girard identity (\ref{eq.thm.symp.newton.ep}) is
similar: a bit easier since it does not require Proposition
\ref{prop.fps.loder} \textbf{(c)}, but a bit messier since signs appear. Note
that all three Newton--Girard identities can also be proved combinatorially or
by induction (this is a homework problem).

The Newton--Girard identities can be used to express the $e$'s, the $h$'s and
the $p$'s in terms of each other; you just need to solve the identities
recursively for each respective sequence. The only wrinkle is that to express
the $e$'s or the $h$'s in terms of the $p$'s, you need to divide by positive
integers, i.e., you need $K$ to be a $\mathbb{Q}$-algebra.

\begin{example}
\label{exa.symp.newton-conversions} \ 

\begin{enumerate}
\item[\textbf{(a)}] Let us write the $p$'s in terms of the $e$'s. The
Newton--Girard identity%
\[
e_{n-1}p_{1}-e_{n-2}p_{2}+e_{n-3}p_{3}-e_{n-4}p_{4}\pm\cdots+\left(
-1\right)  ^{n-1}\underbrace{e_{0}}_{=1}p_{n}=ne_{n}%
\]
can be solved for $p_{n}$ to give%
\begin{align*}
\left(  -1\right)  ^{n-1}p_{n}  &  =ne_{n}-e_{n-1}p_{1}+e_{n-2}p_{2}%
-e_{n-3}p_{3}\\
&  \ \ \ \ \ \ \ \ \ \ +e_{n-4}p_{4}\pm\cdots-\left(  -1\right)  ^{n-2}%
e_{1}p_{n-1}.
\end{align*}
Thus, if we know $e_{1},e_{2},\ldots,e_{n}$ and $p_{1},p_{2},\ldots,p_{n-1}$,
then this formula gives us $p_{n}$. Thus we find, recursively,
\[
p_{1}=e_{1};
\]
then%
\[
-p_{2}=2e_{2}-e_{1}p_{1}=2e_{2}-e_{1}e_{1}=2e_{2}-e_{1}^{2},
\]
so that%
\[
p_{2}=e_{1}^{2}-2e_{2};
\]
then%
\begin{align*}
p_{3}  &  =3e_{3}-e_{2}p_{1}+e_{1}p_{2}\\
&  =3e_{3}-e_{2}e_{1}+e_{1}\left(  e_{1}^{2}-2e_{2}\right) \\
&  =e_{1}^{3}-3e_{1}e_{2}+3e_{3};
\end{align*}
and likewise%
\[
p_{4}=e_{1}^{4}-4e_{1}^{2}e_{2}+2e_{2}^{2}+4e_{1}e_{3}-4e_{4}.
\]
Thus, by recursion, we find a formula for each $p_{n}$ in terms of
$e_{1},e_{2},\ldots,e_{n}$. Note that $e_{i}=0$ for all $i>N$; thus, we will
not have to deal with an ever-growing number of $e$'s as $n$ increases.

\item[\textbf{(b)}] Similarly, we can express the $p$'s in terms of the $h$'s:%
\begin{align*}
p_{1}  &  =h_{1};\\
p_{2}  &  =-h_{1}^{2}+2h_{2};\\
p_{3}  &  =h_{1}^{3}-3h_{1}h_{2}+3h_{3};\\
p_{4}  &  =-h_{1}^{4}+4h_{1}^{2}h_{2}-2h_{2}^{2}-4h_{1}h_{3}+4h_{4}.
\end{align*}
These are the same formulas as for the $p$'s in terms of the $e$'s, but with
different signs! We will soon see that this is no coincidence.

Note, however, that this algorithm expresses each $p_{n}$ in terms of
$h_{1},h_{2},\ldots,h_{n}$, not in terms of $h_{1},h_{2},\ldots,h_{N}$. And
since $h_{i}$ does not become $0$ when $i>N$, the formulas will keep
increasing in complexity with rising $n$.

\item[\textbf{(c)}] Assume that $K$ is a $\mathbb{Q}$-algebra. Then,
recursively applying the second Newton--Girard formula in the opposite
direction, we find%
\begin{align*}
e_{1}  &  =p_{1};\\
e_{2}  &  =\dfrac{1}{2}p_{1}^{2}-\dfrac{1}{2}p_{2};\\
e_{3}  &  =\dfrac{1}{6}p_{1}^{3}-\dfrac{1}{2}p_{1}p_{2}+\dfrac{1}{3}p_{3};\\
e_{4}  &  =\dfrac{1}{24}p_{1}^{4}-\dfrac{1}{4}p_{1}^{2}p_{2}+\dfrac{1}{8}%
p_{2}^{2}+\dfrac{1}{3}p_{1}p_{3}-\dfrac{1}{4}p_{4},
\end{align*}
and so on.
\end{enumerate}

All the coefficients in the above expressions have combinatorial meanings,
which we will see later.
\end{example}

The following theorem -- at least its part \textbf{(a)} -- was originally
proved by Gauss, and has since become fundamental in many places in algebra.
We will prove it in the next lecture.

\begin{theorem}
[Fundamental Theorem of Symmetric Polynomials]\label{thm.symp.fundamental} \ 

\begin{enumerate}
\item[\textbf{(a)}] The elementary symmetric polynomials $e_{1},e_{2}%
,\ldots,e_{N}$ are algebraically independent (over $K$) and generate the
$K$-algebra $\mathcal{S}$. In other words, each $f\in\mathcal{S}$ can be
uniquely written as a polynomial in $e_{1},e_{2},\ldots,e_{N}$. In yet other
words, the map%
\begin{align*}
\underbrace{K\left[  y_{1},y_{2},\ldots,y_{N}\right]  }_{\substack{\text{a
polynomial ring}\\\text{in }N\text{ variables}}}  &  \rightarrow\mathcal{S},\\
g  &  \mapsto g\left(  e_{1},e_{2},\ldots,e_{N}\right)
\end{align*}
is a $K$-algebra isomorphism.

\item[\textbf{(b)}] The complete homogeneous symmetric polynomials
$h_{1},h_{2},\ldots,h_{N}$ are algebraically independent (over $K$) and
generate the $K$-algebra $\mathcal{S}$. In other words, each $f\in\mathcal{S}$
can be uniquely written as a polynomial in $h_{1},h_{2},\ldots,h_{N}$. In yet
other words, the map%
\begin{align*}
\underbrace{K\left[  y_{1},y_{2},\ldots,y_{N}\right]  }_{\substack{\text{a
polynomial ring}\\\text{in }N\text{ variables}}}  &  \rightarrow\mathcal{S},\\
g  &  \mapsto g\left(  h_{1},h_{2},\ldots,h_{N}\right)
\end{align*}
is a $K$-algebra isomorphism.

\item[\textbf{(c)}] Assume that $K$ is a $\mathbb{Q}$-algebra. Then, the power
sums $p_{1},p_{2},\ldots,p_{N}$ are algebraically independent (over $K$) and
generate the $K$-algebra $\mathcal{S}$. In other words, each $f\in\mathcal{S}$
can be uniquely written as a polynomial in $p_{1},p_{2},\ldots,p_{N}$. In yet
other words, the map%
\begin{align*}
\underbrace{K\left[  y_{1},y_{2},\ldots,y_{N}\right]  }_{\substack{\text{a
polynomial ring}\\\text{in }N\text{ variables}}}  &  \rightarrow\mathcal{S},\\
g  &  \mapsto g\left(  p_{1},p_{2},\ldots,p_{N}\right)
\end{align*}
is a $K$-algebra isomorphism.
\end{enumerate}
\end{theorem}

\begin{example}
\label{exa.symp.fundamental} Theorem \ref{thm.symp.fundamental} \textbf{(b)}
tells us that each $h_{n}$ with $n>N$ is a polynomial in the $h_{1}%
,h_{2},\ldots,h_{N}$. For example, for $N=3$, we have%
\[
h_{4}=h_{1}^{4}-3h_{1}^{2}h_{2}+h_{2}^{2}+2h_{1}h_{3}.
\]


Theorem \ref{thm.symp.fundamental} \textbf{(a)} tells us that every symmetric
polynomial in three variables $x,y,z$ can be written as a polynomial in
$e_{1},e_{2},e_{3}$. For example,%
\[
\left(  \left(  x-y\right)  \left(  y-z\right)  \left(  z-x\right)  \right)
^{2}=e_{1}^{2}e_{2}^{2}-4e_{2}^{3}-4e_{1}^{3}e_{3}+18e_{1}e_{2}e_{3}%
-27e_{3}^{2}.
\]

\end{example}

As we said, Theorem \ref{thm.symp.fundamental} will be proved later. For now,
let us notice a simple criterion for proving that a polynomial is symmetric:

\begin{lemma}
\label{lem.symp.simple-transpositions} For each $i\in\left[  N-1\right]  $, we
let $s_{i}$ be the simple transposition in the symmetric group $S_{N}$ that
swaps $i$ with $i+1$ (and leaves all remaining elements of $\left[  N\right]
$ unchanged). Let $f\in\mathcal{P}$. Assume that%
\begin{equation}
s_{k}\cdot f=f\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left[  N-1\right]  .
\label{eq.lem.symp.simple-transpositions.ass}%
\end{equation}
Then, $f$ is symmetric.
\end{lemma}

\begin{proof}
A classical result in permutation combinatorics (see, e.g., \cite[Theorem
5.3.17]{21s}) says that any permutation in $S_{N}$ can be written as a
composition of simple transpositions (i.e., of some of $s_{1},s_{2}%
,\ldots,s_{N-1}$). Thus, for any $\sigma\in S_{N}$, we can write $\sigma$ as
$\sigma=s_{i_{1}}s_{i_{2}}\cdots s_{i_{p-2}}s_{i_{p-1}}s_{i_{p}}$, and then we
have%
\begin{align*}
\sigma\cdot f  &  =\left(  s_{i_{1}}s_{i_{2}}\cdots s_{i_{p-2}}s_{i_{p-1}%
}s_{i_{p}}\right)  \cdot f\\
&  =s_{i_{1}}\cdot s_{i_{2}}\cdot\cdots\cdot s_{i_{p-2}}\cdot s_{i_{p-1}}%
\cdot\underbrace{s_{i_{p}}\cdot f}_{\substack{=f\\\text{(by
(\ref{eq.lem.symp.simple-transpositions.ass}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since we have a group
action of }S_{N}\text{ on }\mathcal{P}\right) \\
&  =s_{i_{1}}\cdot s_{i_{2}}\cdot\cdots\cdot s_{i_{p-2}}\cdot
\underbrace{s_{i_{p-1}}\cdot f}_{\substack{=f\\\text{(by
(\ref{eq.lem.symp.simple-transpositions.ass}))}}}\\
&  =s_{i_{1}}\cdot s_{i_{2}}\cdot\cdots\cdot s_{i_{p-2}}\cdot f\\
&  =\cdots\\
&  =f.
\end{align*}
Thus, $f$ is symmetric.
\end{proof}

\begin{sloppypar}
For example, a polynomial $f\in K\left[  x,y,z\right]  $ is symmetric if and
only if $f\left(  y,x,z\right)  =f$ and $f\left(  x,z,y\right)  =f$.
\end{sloppypar}

\begin{thebibliography}{999}                                                                                              %


\bibitem[21s]{21s}Darij Grinberg, \textit{An Introduction to Algebraic
Combinatorics (Math 531, Winter 2024 lecture notes)}, 22 July 2026.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/t/21s/lecs.pdf}
\end{thebibliography}


\end{document}