\section{The saturated ring \texorpdfstring{$\rr$}{R}}
\subsection{Theory}
\subsubsection{Definitions and first properties}
\begin{Definition*} On the $\lambda$-ring $R_\KK(G)$, define the \textit{saturated} filtration $\lbrace F^n \rbrace_{n}$ as follows:
\[
F^n(G) = \sum_{H\leq G} \ind_H^G(\Gamma^n(H))
\]
\end{Definition*}
This means that $F^n(G)$ is generated by elements of the form:
\[
x = \ind_{H}^G(\gamma^{i_1}(\rho_{1})\cdots \gamma^{i_m}(\rho_{m})), \ \ \ i_1+\cdots+i_m \geq n
\]
with each $\rho_l$ an irreducible representation of $H$. Let $w(x) = i_1+\cdots + i_m$ be the \textit{weight} of $x$.
\begin{Lemma}
\begin{enumerate}
\item Induction and restriction of representations preserve the filtration $F$.
\item $F^i(G)\cdot F^j(G) \subseteq F^{i+j}(G)$.
\item $F^0(G) = R(G), \ F^1(G) = I$, where $I$ is the augmentation ideal.
\end{enumerate}
\end{Lemma}
\begin{proof}
\begin{enumerate}
\item Obviously induction preserves the filtration. For restriction, let $x\in F^i(G)$. It is enough to prove that if $x = \ind_H^G(y)$ with $y \in \Gamma^i(H)$ then $\res_K(x) \in F^i(K)$ for all $k \leq G$. For this, we use the double coset formula: let $S$ be a set of $(H,K)$-double coset representatives of $G$. For $s\in S$, let
\[
H_s = sHs^{-1}\cap K \leq K
\]
and let
\[
y^s(g) = y(s^{-1}gs), \ \ \ \text{for all} \ \ \ g \in H_s.
\]
Then $y^s$ is a representation of $H_s$ and
\[
\res_K\ind_H^G(y) = \sum_{s\in K \backslash G / H} \ind_{H_s}^K(y^s)
\]
Note that each $y^s$ is in $\Gamma^i(H_s)$, since exterior powers commute with conjugation (so the $\lambda$-operations commute with the operation $y \mapsto y^s$). Thus $\res_K(x) \in F^i(K)$.
\item It is sufficient to prove that if $\tilde{x} = \ind_H^G(x)$ and $\tilde{y} = \ind_K^G(y)$ with $x\in \Gamma^i(H)$ and $y \in \Gamma^j(K)$ then $\tilde{x}\cdot\tilde{y} \in F^{i+j}(G)$. We proceed by induction on the order of $G$. Suppose $H < G$ is a proper subgroup (otherwise there is nothing to prove). By the projection formula:
\[
\tilde{x}\tilde{y} = \ind_H^G(x)\ind_K^G(y) = \ind_H^G(x\res_H\ind_K^G(y)).
\]
Since restriction preserves the filtration, $\res_H\ind_K^G(y) \in F^j(H)$, so:
\[
x\cdot\res_H\ind_K^G(y) \in F^i(H)F^j(H) \subseteq F^{i+j}(H).
\]
where the inclusion is true by induction. In conclusion:
\[
\ind_H^G(x\cdot\res_H\ind_K^G(y)) \in F^{i+j}(G).
\]
\item The fact that $F^0(G) = R(G)$ comes from the fact that $\Gamma^0(G) = R(G)$. For $F^1(G)$, simply observe that, since $\varepsilon(\ind_H^G(\rho)) = [G:H]\varepsilon(\rho)$:
\[
F^1(G) = \sum_{H \leq G} \ind_H^G(\ker(\varepsilon\vert_H)) = \ker(\varepsilon) = I.
\]
\end{enumerate}
\end{proof}
\begin{Definition*}The graded ring
\[
\rr_\KK(G) = \bigoplus_{i\geq 0}F^i(G)/F^{i+1}(G).
\]
is called the \textit{saturated} (graded representation) ring associated to $G$. In the sequel, we drop the subscript whenever $\KK$ is clear from the context.
\end{Definition*}
\begin{Theorem}
$\rr^*$ is a Mackey functor, with
\begin{align*}
\ind_K^H: \rr^*(K) &\longrightarrow \rr^*(H) \\
\res_K^H: \rr^*(H) &\longrightarrow \rr^*(K) \\
c_g: \rr^*(H) &\longrightarrow \rr^*(^gH)
\end{align*}
\end{Theorem}
\begin{proof}
The only nontrivial statement is that conjugation of representations gives rise to a well-defined homomorphism $c_g: \rr^*(H) \rightarrow \rr^*(H^g)$. The rest of the axioms follow from basic representation theory. So, consider an element of $F^i(H)$ of the form $\ind_K^H(C_{i_1}(\rho_1)\cdots C_{i_n}(\rho_n))$ for some representations $\rho_l \in R(K)$ and $K\leq H$. Note that since exterior powers commute with conjugation, one has $c_g(C_i(\rho)) = C_i(c_g(\rho))$ (in other words, $c_g$ is compatible with the $\Gamma$-filtration). Then
\[
c_g\ind_K^H(C_{i_1}(\rho_1)\cdots C_{i_n}(\rho_n)) = \ind_{^gK}^{^gH}c_g(C_{i_1}(\rho_1)\cdots C_{i_n}(\rho_n)) = \ind_{^gK}^{^gH}(C_{i_1}(c_g(\rho_1))\cdots C_{i_n}(c_g(\rho_n)))
\]
and $\ind_{^gK}^{^gH}(C_{i_1}(c_g(\rho_1))\cdots C_{i_n}(c_g(\rho_n))) \in F^i(^gH)$ as required.
\end{proof}
Note that $\rr^*$ is actually a Green functor, that is, a Mackey functor with an $R$-algebra structure compatible with restriction and satisfying the projection formula.
Since $\Gamma^n \subseteq F^n$ for all $n \geq 0$, there is a natural map of graded rings:
\[
\eta: R^*(G) \longrightarrow \rr^*(G)
\]
induced by the identity.
\begin{Definition*}
We say that $R^*_\KK(G)$ is \textit{saturated} if the natural map $\eta$ is an isomorphism. A group $G$ is \textit{saturated (over $\KK$)} if $R^*_\KK(G)$ is saturated. For $H \leq G$, if the induction $i_*:R(H) \rightarrow R(G)$ is compatible with the filtration $(\Gamma^n)$, then $H$ is \textit{$\Gamma$-compatible} with $G$.
\end{Definition*}
Note that by definition, the saturated ring $\rr^*(G)$ is generated by Chern classes of irreducible characters $G$, as well as classes of the form $\ind_{H}^{G}(c_i(\rho))$ with $\rho$ a virtual character of $H \leq G$. Denote
\[
d_i(\rho) := c_i^{H}(\rho) := \ind_{H}^{G}(c_i(\rho))
\]
Below are some conditions under which $G$ is saturated:
\begin{Lemma}
If the restriction maps $i^*: R(G) \rightarrow R(H)$ are surjective for all $H\leq G$, then $G$ is saturated.
\end{Lemma}
\begin{proof}
Let $\rho$ be a virtual representation of $G$, with $\rho$ in $\Gamma^M(G)$ say. Then:
\[
i_*(\rho) = i_*i^*(\sigma) = i_*(\mathbb{1})\rho \in \Gamma^n(G).
\]
So all virtual representations in $F^n(G)$ (which are induced from subgroups of $G$) are also in the $\Gamma^n(G)$, and thus $\rr^*(G) = R^*(G)$.
\end{proof}
Note however that the converse isn't true: in the case of $D_p$, restriction of representations to $C_p$ isn't surjective, but the induction still preserves the filtration.
\begin{Proposition}
The filtrations $(F^n)_n$ and $(\Gamma^n)_n$ induce the same topology on $R(G)$.
\end{Proposition}
\begin{proof}
Let $U\subseteq R(G)$ be open for the $F$-topology. Then by definition, for any $x\in U$ there is an integer $N$ such that $x+F^N \subset U$. Since $\Gamma^N \subseteq F^N$, we also have $x+\Gamma^N \subseteq U$. Thus $U$ is also open for the $\Gamma$-topology.\\
To prove that a set $U$ open in the $\Gamma$-topology is also open in the $F$-topology, we need to show that for each $N$, there is an $M$ such that $F^M \subseteq \Gamma^N$. We use \cite[Th. 6.1]{atiyah-characters_cohomology}: let $H \leq G$, and recall that $R(H)$ can be viewed as an $R(G)$-module via the restriction homomorphism. Then the $I(H)$-adic topology is equal to the $I(G)$-adic topology. By \cref{gammatopology_coincides_idaictopology}, these topologies are also equal to the $\Gamma$-topology on $H$. In particular, for some $k$,$m$ we have
\[
\Gamma^N(H) \supset I(G)^k\cdot R(H) \supset \Gamma^m(H)
\]
Pick $k,l$ large enough that we also have $I(G)^k \subset \Gamma^N(G)$. Then
\[
\ind_H^G(\Gamma^m(H)) \subset \ind_H^G(I(G)^k\cdot R(H)) \subset I(G)^k \subset \Gamma^N(G)
\]
Now let
\[
M = \max_{H\leq G} \ \left\lbrace \min \ \{m \ \vert \ \ind_H^G(\Gamma^m(H)) \subset \Gamma^N(G)\}\right\rbrace,
\]
then $F^M(G) = \sum_{H\leq G}\ind_H^G(\Gamma^M(H)) \subset \Gamma^N(G)$.
\end{proof}
\begin{Corollary}
If the natural map $\eta: R(G) \rightarrow \rr^*(G)$ is surjective, then it is an isomorphism and the filtrations $(F^n)$ and $(\Gamma^n)$ are equal.
\end{Corollary}
\begin{proof}
If $\eta$ is surjective, then $\rr^*(G)$ is generated by Chern classes of elements of $R(G)$. Let $P_{w}$ denote a polynomial of total weight $w$, then any $x\in F^n(G)$ can be written as
\begin{align*}
x &= P_{n}(C_{i_1}(\rho_{1}), \cdots, C_{i_k}(\rho_k)) + y_{n+1} \\
&= P_{n}(C_{i_1}(\rho_{1}), \cdots, C_{i_k}(\rho_k)) + P_{n+1}(C_{i_1}(\rho_{1}), \cdots, C_{i_k}(\rho_k)) + y_{n+2} \\
&= \sum_{l = 1}^m P_{n+l}(C_{i_1}(\rho_{1}), \cdots, C_{i_k}(\rho_k)) + y_{m+1}
\end{align*}
where the $\rho_j$'s are irreducible representations of $G$ and $y_t \in F^{t}$. So $x$ is in $\Gamma^n(G) + F^m(G)$ for all $m$, that is, $x$ is in the topological closure $\overline{\Gamma^n(G)}$ of $\Gamma^n(G)$. But $\Gamma^n(G)$ is a closed set in $R(G)$, thus $x \in \Gamma^n(G)$.
\end{proof}
Finally, we can use transitivity of induction and restriction: suppose $\sigma \in \Gamma^n(H)$ with $H\leq K \leq G$, and $R^*(K)$ is saturated. Then $\ind_H^K(\sigma) \in \Gamma^n(K)$, thus we do not need to restrict representations to $H$, but only to $K$.
\subsubsection{Stable elements, Swan's lemma}
Consider a sugroup $H \leq G$.
\begin{Definition*}
An element $x \in \rr^*(H)$ is called \textit{stable} if
\[
\res^{^gH}_{^gH\cap H}(c_g(x)) = \res^H_{^gH\cap H}(x)
\]
for all $g\in G$.
\end{Definition*}
We want to prove:
\begin{Theorem} \label{theorem_swan}
Let $G \geq H \geq Syl_p(G)$ where $Syl_p(G)$ is a $p$-Sylow of $G$ and let $\rr^*(G)_{(p)}$ denote the $p$-primary component of $\rr^*(G)$. Then:
\[
\res_H^G : \rr^*(G)_{(p)} \longrightarrow \rr^*(H)_{(p)}.
\]
is injective, and its image consists of the stable elements in $\rr^*(H)_{(p)}$
\end{Theorem}
\begin{proof} \textit{Step 1.} $\res_H^G$ is injective on $p$-primary components.\\
Let $x \in F^n(G)$, then $\ind_H^G\res_H^G(x) = \KK[G/H] \ (\Mod F^{n+1}(G))$. But
\[
\KK[G/H] = \KK[G/H] - (\KK[G/H] - \varepsilon(\KK[G/H])) = \varepsilon(\KK[G/H]) = [G:H] \ (\Mod F^{1}(G))
\]
so that
\[
\ind_H^G\res_H^G(x) = [G:H]x \ (\Mod F^{n+1}(G))
\]
which is injective on $p$-primary components since $H$ contains a $p$-Sylow of $G$. Thus $\res_H^G$ is injective on $p$-primary components, as needed.\\
\textit{Step 2.} If $x \in \Ima(\res_H^G)$ then $x$ is stable. \\
Write $x = \res_H^G(y)$. Then, since $y$ is a character of $G$ we have $c_g(y) = y$ for all $g\in G$ and:
\begin{align*}
\res^{^gH}_{^gH\cap H}(c_g(x)) &= \res^{^gH}_{^gH\cap H}(c_g\res_H^G(y)) = \res^{^gH}_{^gH\cap H}(\res_{^gH}^Gc_g(y)) = \res^{^gH}_{^gH\cap H}(\res_{^gH}^G(y)) \\
&= \res_{^gH\cap H}^G(x) = \res_{^gH\cap H}^H\res_H^G(y) \\
&= \res_{^gH\cap H}^H(x)
\end{align*}
\textit{Step 3.} If $x \in \rr^*(H)_{(p)}$ is stable then $x \in \res_H^G(\rr^*(G)_{(p)})$. \\
First observe that if $x$ is stable, then by the double coset formula:
\begin{align*}
\res_H^G\ind_H^G(x) &= \sum_{g \in [H\backslash G /H]} \ind_{^gH\cap H}^H(x^g) = \sum_{g \in [H\backslash G /H]} \ind_{^gH\cap H}^H\res^{^gH}_{^gH\cap H}c_g(x) \\
&= \sum_{g \in [H\backslash G /H]} \ind_{^gH\cap H}^H\res_{^gH\cap H}^H(x) = \sum_{g \in [H\backslash G /H]} \KK[H/^gH\cap H]x.
\end{align*}
Plugging $x = \mathbb{1}$ into this equality, we get:
\[
\res_H^G(\KK[G/H]) = \res_H^G\ind_H^G(\mathbb{1}) = \sum_{g \in [H\backslash G /H]}\KK[H/^gH\cap H]
\]
so that whenever $x$ is stable, we have $\res_H^G\ind_H^G(x) = \res_H^G(\KK[G/H])x = [G:H]x$. Now, say $\vert Syl_p(G) \vert = p^i$, and choose an integer $q$ so that $q[G:H] = 1 \ (\Mod p^i)$. Then
\[
\res_H^G\ind_H^G(qx) = q[G:H]x = x \ (\Mod p^i)
\]
which implies that $x \in \res_H^G(\rr^*(G)_{(p)})$ whenever $x\in \rr^*(H)_{(p)}$ is stable.
\end{proof}
\begin{Corollary}
If $H \unlhd G$ is a normal subgroup such that $H \supseteq Syl_p(G)$, then
\[
\rr^*(G)_{(p)} \cong \Ima(\res_H^G) = \rr^*(H)^{G/H}_{(p)}
\]
\end{Corollary}
\begin{proof}
If $H$ is normal, the stability condition becomes $c_g(x) = x$, that is, $x$ is invariant by the action of $G/H$.
\end{proof}
\begin{Corollary} \label{swanslemma}
If $H := Syl_p(G)$ is abelian, then
\[
\res_H^G: \rr^*(G)_{(p)} \longrightarrow \rr^*(H)^{N_G(H)}
\]
is an isomorphism.
\end{Corollary}
\begin{proof}
By \cref{theorem_swan}, $\res_H^G$ is an isomorphism on stable elements in $\rr^*(H)_{(p)}$. Moreover, $\rr^*(H)$ is $p^i$-torsion for some $i$, so that $\res_H^G$ is in fact surjective on all stable elements of $\rr^*(H)$. Obviously any stable element is in particular invariant under the action of $N_G(H)$, so we only need to show that all elements of $\rr^*(H)^{N_G(H)}$ are stable. \\
Let $g \in G$ and let $C$ be the centralizer of $^gH\cap H$ in $G$. Since $H$ is abelian, $H, \ ^gH \subseteq C$, and $H$ and $^gH$ are $p$-Sylows of $C$. Thus for some $t \in C$ we have $^{tg}H = H$. Then
\begin{align*}
\res^{H}_{^gH\cap H}c_{tg}(x) &= \res^{H}_{^gH\cap H}c_tc_g(x) \\
&= c_t\left(\res^{H^t}_{(^gH\cap H)^t}c_g(x)\right) \\
&= \res^{H^t}_{^gH\cap H}c_g(x) \ \ \ \ \text{since $t \in C$}\\
&= \res^{^gH}_{^gH\cap H}c_g(x) \ \ \ \ \text{since $H^t = {}^gH$}\\
\end{align*}
Now if $x \in \rr^*(H)^{N_G(H)}$ then $c_{tg}(x) = x$ thus:
\[
\res^{^gH}_{^gH\cap H}c_g(x) = \res_{^gH\cap H}^H(x)
\]
and $x$ is stable.
\end{proof}
\subsection{Examples}
\subsubsection{Groups that are saturated}
\begin{Proposition}
Abelian groups are saturated over $\CC$.
\end{Proposition}
\begin{proof}
Let $G$ be an abelian group and define $\widehat{G} := \Hom(G,\CC^*)$. Then any group homomorphism (of abelian groups) $\phi: G \rightarrow H$ induces a map $\widehat{\phi} : \widehat{H} \rightarrow \widehat{G}$, which is injective if, and only if, $\phi$ is surjective. Additionally, there is a natural isomorphism between $G$ and its double dual $\widehat{\widehat{G}}$ given by associating to $g$ the evaluation in $g$.\\
Now if $H\leq G$, then the injection $H \rightarrow G$ induces a map $\widehat{\phi}: \widehat{G} \rightarrow \widehat{H}$, and also a map $\widehat{\widehat{\phi}}:\widehat{\widehat{H}} \rightarrow \widehat{\widehat{G}}$. The latter map is injective, which means by the above that $\widehat{\phi}$ is surjective. Thus the characters of $H$ all come from restrictions of characters of $G$, and $G$ is saturated.
\end{proof}
\begin{Proposition} \label{Q8_is_saturated}
The group $Q_8$ is saturated.
\end{Proposition}
\begin{proof}
The quaternion group contains one subgroup isomorphic to $C_2$, which is generated by $-1$, and three subgroups isomorphic to $C_4$, which all contain $C_2$ and are all conjugate, generated respectively by $i$, $j$ and $k$. Since all these groups are saturated, we only need to check that the maximal saturated subgroup $H = \langle k \rangle \cong C_4$ is $\Gamma$-compatible with $Q_8$, which we do manually.\\
Recall that
\[
R^*(Q_8) = \frac{\ZZ[x_1,x_2,y]}{(2x_i, 8y, x_i^2, x_1x_2 - 4y)}, \ \ \ \ R^*(C_4) = \frac{\ZZ[t]}{(4t)}
\]
where, in particular, $y = c_2(\Delta)$, with $\Delta$ the irreducible representation of $Q_8$ of degree 2, and $t = c_1(\rho)$ with $\rho$ the usual generator of $R(C_4)$. Note that $i_*(C_1(\rho)) \in \Gamma^1(Q_8) = I_{Q_8}$. Moreover, the representation $\Delta$ restricts on $C_4$ to $\rho + \rho^{-1}$,and so
\[
\res^{Q_8}_{C_4}(y) = c_2(\rho+\rho^{^-1}) = -c_1(\rho)^2 = -t^2.
\]
therefore $C_1(\rho)^2 = \res(-Y)$, and so
\[
i_*(C_1(\rho)^2) = i_*(\res(Y)) = \CC[Q_8/C_4].Y \in \Gamma^2(Q_8).
\]
Thus:
\[
i_*(C_1(\rho)^n) = i_*(C_1(\rho)^{2m+l}) = i_*(C_1(\rho))i_*(Y^m) \in \Gamma^l\cdot\Gamma^{2m} \ \text{for} \ l = 0,1
\]
which means that $C_4$ is $\Gamma$-compatible with $Q_8$, and therefore $Q_8$ is saturated.
\end{proof}
\begin{Proposition}
Let $p$ be an odd prime, then the dihedral group $D_p$ of order $2p$ is saturated.
\end{Proposition}
\begin{proof}
Recall that
\[
R^*(G) = \frac{\ZZ\left[x,y\right]}{(2x,py,xy)}.
\]
Since $D_p = C_p \rtimes C_2$ and $C_p$, $C_2$ are abelian, these are the maximal saturated subgroups of $D_p$. The signature $\varepsilon$ of $D_p$ restricts on $C_2$ to the representation $\rho$, which generates $R(C_2)$. Thus $C_2$ is $\Gamma$-compatible with $D_p$, and we only need to look at $C_p$.\\
Since $\res(Y) = -C_1(\rho)^2$ the same argument as in the proof of \cref{Q8_is_saturated} applies. \\
\end{proof}
Another idea is to quantify the obstruction to $i_*$ being a ring homomorphism.
\begin{Lemma}
Suppose $H$ is normal in $G$ and $y \in R(H)$ is stable by $G$, that is:
\[
y(\sigma\tau\sigma^{-1}) = y(\tau)
\]
for any $\tau \in H$, $\sigma \in G$. Then for any $x \in R(H)$
\[
i_*(x)i_*(y) = i_*(xy)i_*(1)
\]
\end{Lemma}
\begin{proof}
If $y$ is stable by $G$ then the induced character $i_*(y)$ is $i_*(1)y'$, where the values of $y'$ are those of $y$ on $H$ and 0 on $G\setminus H$. In other words we have $\res (y') = y$. Thus:
\[
i_*(x)i_*(y) = i_*(x)i_*(1)y' = i_*(x \res(y'))i_*(1) = i_*(xy)i_*(1).
\]
\end{proof}
\subsubsection{The saturated ring of \texorpdfstring{$A_4$}{A4}}
\begin{Proposition} \label{saturatedring_A4}
\[
\rr^*(A_4) = \frac{\ZZ[x,y,z]}{(3x,2y,2z,y^3-z^2)} \cong \frac{\ZZ[x,y',z]}{(3x,6y',2z,2y'-x^2, x^6 + y'^3 - z^2)}
\]
with $x = c_1(\rho), y' = c_2(\theta), z = d_3(\rho_{(1,0)}-\rho_{(0,1)})$ where $\rho_{(i,j)}$ are representations of $C_2\times C_2$.
\end{Proposition}
\begin{proof}
The group $A_4$ has the following $p$-Sylow structure:
\begin{enumerate}
\item For $p=2$, we have $Syl_2(A_4) \cong C_2\times C_2$, generated by $(12)(34)$ and $(13)(24)$. The group $C_2\times C_2$ is normal in $A_4$.
\item For $p=3$, we have $Syl_3(A_4) \cong C_3$, generated by say $(123)$; it is self-normalizing.
\end{enumerate}
\textit{$3$-primary part.} Since $C_3$ is self-normalizing and abelian, the action of its normalizer is trivial. Thus
\[
\rr^*(A_4)_{(3)} \cong \rr^*(C_3) = \frac{\ZZ[t]}{(3t)}
\]
\textit{$2$-primary part.} Elements of $A_4$ act on $C_2\times C_2$ by circular permutations of the three nontrivial elements. Since $C_2\times C_2$ is abelian, it is saturated so:
\[
\rr^*(C_2\times C_2) \cong \frac{\ZZ[t_1,t_2]}{(2t_1,2t_2, t_1^2t_2 - t_1t_2^2)}
\]
the stable elements are those invariant under circular permutations of $t_1, t_2, t_1+t_2$. There are no such elements in degree $1$. For $n \geq 2$, we have $\rr^n(C_2\times C_2) = \langle t_1^n, t_2^n, t_1^{n-1}t_2 \rangle$, and the action of $A_4$ is described by the permutation matrix:
\[
\begin{pmatrix} 0 & 1 & 0 \\
1 & 1 & 1 \\
0 & 0 & 1
\end{pmatrix}
\]
whose (one-dimensional) invariant subspace is $t_1^n + t_2^n + t_1^{n-1}t_2$. Let $y = t_1^2 + t_1t_2 + t_2^2$ and $z = t_1^3 + t_1^2t_2 + t_2^3$, then
\[
\rr^*(A_4)_{(2)} \cong \rr^*(C_2\times C_2)^{A_4} \cong \frac{\ZZ[y,z]}{(2y,2z, y^3 - z^2)}.
\]
Define $y' = y - x^2$ to obtain the presentation
\[
\rr^*(A_4) = \frac{\ZZ[x,y',z]}{(3x,6y',2z,2y'-x^2, x^6 + y'^3 - z^2)}.
\]
\textit{Explicit generators.} We determine the images of the classes $c_1(\rho), c_2(\theta)$ under the natural map $\eta$. For this, we use that the maps $\res$ and $\eta$ commute.\\
\begin{itemize}
\item The class $c_1(\rho)$ restricts to the degree one generator $t$ of $\rr^*(C_3)$ and to $0$ on $C_2\times C_2$. Thus its $3$-primary part must be equal to $x$ and its $2$-primary part to $0$, and $[c_1(\rho)] = x$.
\item The class $c_2(\theta)$ restricts to $-t^2$ on $C_3$ and to $t_1^2 + t_1t_2 + t^2_2$ on $C_2\times C_2$, thus its $3$-torsion part is $-x^2$ and its $2$-primary part is $y$, so $[c_2(\theta)] = y - x^2 = y'$.
\item For $z$: write $i_* = \ind_{C_2\times C_2}^{A_4}$, $i^* = \res^{A_4}_{C_2\times C_2}$, and consider:
\begin{align*}
i^*(z) &= t_1^3 + t_1^2t_2 + t_2^3 \\
i_*i^*(z) &= i_*(t_1^3+t_1^2t_2+t_2^3) \\
3z = z &= i_*(t_1^3+t_1^2t_2+t_2^3)
\end{align*}
and note that $t_1(t_1^2+t_1t_2+t_2^2) = t_1^3$. Then
\begin{align*}
i_*(t_1^3) &= i_*(t_1(t_1^2 + t_1t_2+t_2^2)) = i_*(t_1\cdot i^*(c_2(\theta))) \\
&= i_*(t_1)c_2(\theta)
\end{align*}
Now $2t_1 = 0$ so $2i_*(t_1) = 0$, but $\rr^1(G)$ is $3$-torsion, thus $i_*(t_1) = 0$. This means that $i_*(t_1^3) = 0$ and
\[
z = i_*(t_1^2t_2+t_2^3)
\]
a direct computation then shows that $c_3(\rho_{(1,0)}-\rho_{(0,1)}) = t_1^2t_2+t_2^3$, so that $z = i_*(\rho_{(1,0)}-\rho_{(0,1)}) = d_3(\rho_{(1,0)}-\rho_{(0,1)})$.
\end{itemize}
\end{proof}
\subsubsection{The saturated ring of \texorpdfstring{$PSL(2,p)$}{PSL(2,p)}}
Let $G = PSL(2,p)$ be the projective special linear group over $\FF_p$, where $p$ is an odd prime such that $p \equiv 3,5 (\Mod 8)$. The order of $G$ is $|G| = \frac{p(p+1)(p-1)}{2}$
\paragraph{Sylow structure and stable elements.} For each prime $l$ dividing the order of $G$, let $H_l = Syl_l(G)$ and $N_l = N_G(H_l)$. There are $4$ possible cases:
\begin{enumerate}
\item $l = p$. Then $H_p \cong C_p$ is generated by the matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The normalizer of $H_p$ is the group:
\[
N_p = \left\lbrace \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \in PSL(2,p) \right\rbrace,
\]
with action
\[
\begin{pmatrix} a & b \\ 0 & a^{-1}\end{pmatrix} \cdot \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} a^{-1} & -b \\ 0 & a \end{pmatrix} = \begin{pmatrix} 1 & a^2n \\ 0 & 1 \end{pmatrix},
\]
inducing $\rho \mapsto \rho^{a^2}$ for a generator $\rho$ of $R(C_p)$. On $ \rr^*(C_p) \cong \frac{\ZZ[x]}{(px)} $, this induces $x\mapsto a^2x$. The subring generated by $x^{\frac{p-1}{2}}$ is stable by this action, and conversely if $a$ is an element of multiplicative order $(p-1)$, then a monomial $x^m$ being stable by the action $x \mapsto a^2x$ implies that $m$ is a multiple of $\frac{p-1}{2}$. Thus
\begin{equation}
\rr^*(H_p)^{N_p} \cong \frac{\ZZ[t]}{(pt)}, \ \ \ \ |t| = \frac{p-1}{2}.
\end{equation}
\item $l$ is an odd prime dividing $p-1$. Then $H_l \cong C_{l^i}$ for some integer $i$, generated by $\begin{pmatrix} n & 0 \\ 0 & n^{-1} \end{pmatrix} $ for some $n$ of order $l^i$ in $\FF_p\x$. A straightforward computation gives that $N_l := N^{G}(H_l)$ is generated by diagonal matrices (which commute with the elements of $H_l$) together with the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ which sends an element $h \in H_l$ to its inverse. The induced action on the representation ring is $\rho \mapsto \rho^{-1}$, which translates as $x \mapsto -x$ in the graded ring. Thus
\begin{equation}
\rr^*(H_l)^{N_l} \cong \frac{\ZZ[y]}{(l^iy)}, \ \ \ \ |y| = 2.
\end{equation}
\item $l=r$ is an odd prime dividing $p+1$. We prove that $H_r$ is cyclic. Note that the $r$-Sylow of $G$ is isomorphic to that of $G' := PSL(2,p^2)$ since the index of $G$ in $G'$ is coprime to $r$. Let $\alpha \in \FF_{p^2}\x$ have multiplicative order $r^i$. The matrix $A' = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha^{-1} \end{pmatrix}$ generates a cyclic group $H_r'$ of order $r^i$ in $G'$, which is thus an $r$-Sylow subgroup. Note that $\alpha \notin \FF_p\x$, however any matrix of $G$ similar to $A$ generates an isomorphic group in $G$. One can take for example $A = \begin{pmatrix} 0 & -1 \\ 1 & \alpha+\alpha^{-1} \end{pmatrix}$, the companion matrix to the minimal polynomial of $\alpha$.\\
The normalizer $N_r'$ of $C_{r^i}'$ in $G'$ is a dihedral group of order $p^2-1$, generated by all diagonal matrices together with the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ which sends $A$ to its inverse. The change of basis sending $A$ to $A'$ allows us to view $N_r$ as a subgroup of $N_r'$, and thus the elements of $N_r$ act either trivially or by inversion on $H_r$.\\
It remains to show that there exists a matrix $S$ such that $S^{-1}AS = A^{-1}$. Let $a = \alpha+\alpha^{-1}$. By a direct calculation, one shows that any matrix of the form $\begin{pmatrix} -x & y \\ ax+y & x \end{pmatrix}$ in $GL(2,p)$ satisfies this property, thus $S \in PSL(2,p)$ exists if and only if there is a pair $(x,y) \in \Fp^2$ such that $-x^2 -axy - y^2 = 1$. This equation is equivalent to $X^2 +1 = bY^2$, with~$X = x+\frac{a^2}{4}y$, $Y = y$ and~$b =\frac{a^2}{4} -1$. There are $(p+1)/2$ squares in $\Fp$ (including $0$), so there are $(p+1)/2$ elements of the form $X^2+1$, and, if $b \neq 0$ then there are also $(p+1)/2$ elements of the form $bY^2$. Thus whenever $b \neq 0$, the set of elements of the form $X^2+1$ and the set of elements of the form $bY^2$ have nontrivial intersection, and there is a solution to $x^2 axy + y^2 = -1$. Now, $b = 0$ if and only if $a^2 = 4$, that is, $a = \pm 2 (\Mod p)$. But then $\alpha$ is a solution of $t^2 \pm 2t +1$, that is, $\alpha = \alpha^{-1}$ has multiplicative order $2$, in contradiction with our assumption. Thus $b$ is always nonzero, which completes the proof.\\
We have:
\begin{equation}
\rr^*(H_r)^{N_r} \cong \frac{\ZZ[y]}{(r^iy)}, \ \ \ \ |y| = 2.
\end{equation}
\item $l = 2$. Since $p \equiv 3,5 (\mod 8)$, the $2$-Sylow subgroup of $G$ has order $4$. There are two cases:
\begin{itemize}
\item if $p \equiv 5 (\mod 8)$, let $a$ satisfy $a^2 \equiv -1 (\Mod p)$. Then
\[
H_2 = \left\langle h_1 := \begin{pmatrix} a & 0 \\ 0 & -a \end{pmatrix}, h_2 := \begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} \right\rangle.
\]
I claim that $N_2 \cong A_4$. First, we have $C_G(h_1)\cap N_G(H_2) = \{Id\}$, as a direct calculation shows, and similarly for $h_2$ and $h_1h_2 =: h_3$. Therefore, if $N \in N_2$ acts nontrivially on $H_2$, it must permute all $3$ nontrivial elements. If $T = \begin{pmatrix} x & -ax \\ x & ax \end{pmatrix}$, with $x^2 = \frac{1}{2a}$, then $Th_1T^{-1} = h_2$ and $Th_2T^{-1} = h_3$. Both $2$ and $a$ are nonresidues $\Mod p$ since $p \equiv 5 (\Mod 8)$ and if $a$ were a residue, then $PSL(2,p)$ would contain an element of order 4, contradicting $H_2 \cong C_2\times C_2$. Thus there is an $x$ satisfying $x^2 = 1/2a$. Moreover $T$ is unique up to multiplication by an element of $C_G(H_2) = H_2$, which shows that $N_2 = \left\langle T, H_2 \right\rangle \cong A_4$.\\
\item if $p \equiv 3 (\Mod p)$, let $b$ satisfy $b^2 \equiv -2 (\Mod p)$. Then
\[
H_2 = \left\langle \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} b & 1 \\ 1 & -b \end{pmatrix} \right\rangle
\]
Again, we have $N_2 \cong A_4$ acting by cyclic permutations, generated by $H_2$ together with the matrix $T = \begin{pmatrix} \frac{1}{b} & \frac{1}{b} \\ -\frac{b+2}{2} & \frac{b-2}{2} \end{pmatrix}$.\\
\end{itemize}
In both cases the normalizer acts as cyclic permutations on the nontrivial elements of $H_2$, and thus:
\begin{equation}
\rr^*(H_2)^{N_2} \cong \frac{\ZZ[y,z]}{(2y,2z, y^3 - z^2)}, \ \ \ \ |y| = 2, |z| = 3.
\end{equation}
\end{enumerate}
Putting all of this together, we get:
\begin{Theorem}
Let $G = PSL(2,p)$ be the projective special linear group over $\FF_p$, where $p$ is an odd prime such that $p \equiv 3,5 (\Mod 8)$. Write:
\[
|G| = 4\cdot p\cdot l_1^{i_1}\cdots l_n^{i_n}\cdot r_1^{j_1}\cdots r_m^{j_m}, \ \ \ \ \text{ with } \ \ l_k|(p-1), \ \ r_k|(p+1).
\]
Then:
\begin{equation}
\rr^*(G) \cong \frac{\ZZ[x_1,\cdots,x_n, y_1,\cdots y_m, z, t, u]}{(l_k^{i_k}x_k, r_k^{j_k}y_k, 2z,2t, pu, z^3 - t^2)}
\end{equation}
with $|x_k| = |y_k| = |z| = 2$, $|t| = 3$, $|u| = (p-1)/2$.
\end{Theorem}
