\subsection{Induction as a transfer for $D_p$}
Fix a finite group $G$ and a subgroup $H$ of $G$. Then any (complex) representation $\rho$ of $H$ induces a complex representation $\mathrm{Ind}_{H}^{G}(\rho)$ of $G$. This induction can be extended by linearity to an additive group homomorphism $T: R(H) \rightarrow R(G)$. However, $T$ is not in general a ring homomorphim, since $T(1) = \CC\left[G/H\right]$.
\paragraph*{Question.} \textit{Is $T$ compatible with the graded ring structure on $R(H)$ and $R(G)$?}\\
More precisely, we'd like the following property to be true:
\[
T(\Gamma^N(H)) \subset \Gamma^N(G)
\]
for any $N \in \NN$.
\paragraph*{Useful properties.} \begin{enumerate}
\item Projection formula: $T(x\cdot\res(y)) = T(x)y$
\item Augmentation: $\varepsilon(T(x)) = \left[G:H\right]\varepsilon(x)$, so in particular $T(I_H) \subset I_G$ where $I_H, I_G$ are augmentation ideals.
\end{enumerate}
\begin{Example}
Let $p$ be an odd prime. Consider the dihedral group $G=D_p$ of order $2p$ and its normal subgroup $H = C_p$. Fix $\rho$, an irreducible (nontrivial) representation of $H$, and recall that
\[
R^*(G) = \frac{\ZZ\left[x,y\right]}{(2x,py,xy)}
\]
with $\vert x \vert = 1$, $\vert y \vert = 2$. Let $t = C_1(\rho) \in R(H)$, then $\res(Y) = t^2$ (where $C_1(\rho), Y $ are some lifts of $c_1(\rho), y$). Since $H$ is abelian, we have $\Gamma^N(H) = (\rho-1)^N = (t^N)$.\\
Recall that $R(H)$ is generated by powers of $t$; so we need to look at the effect of $T$ on those. We have:
\begin{itemize}
\item $T(t) \in I_G = \Gamma^1$ as we noticed above.
\item $T(t^2) = T(1\cdot \res(Y)) = T(1)\cdot Y$ by the projection formula, and $Y \in \Gamma^2$ by definition.
\item In general, $T(t^n) = T(t^{2k+i}) = T(t)^iY^k \in \Gamma^i\cdot\Gamma^{2k} \subset \Gamma^{2k+i}$ for $i=0,1$.
\end{itemize}
So in this case we have $T(\Gamma^N(H)) \subset \Gamma^N(G)$.
\begin{Remark} This also works for the other nontrivial subgroup $K=C_2$ of $D_p$, since the generator $z$ of $R^*(K)$ is the restriction of the generator $x$ of degree 1 of $R^*(G)$.
\end{Remark}
\end{Example}
\begin{Remark} Suppose that some element $x \in \Gamma^1(H)$ satisfies $x^k = \res(y)$ for some $y \in R(G)$. Then, proceeding as in the above calculation, we can show that if $T(x^i) \in \Gamma^i$ for all $i = 1,\cdots,k-1$ (and $y\in\Gamma^k$) then for all $N$,
\[
T(x^N)= T(x^{mk+i}) = T(x^i)y^m \in \Gamma^i\Gamma^{mk}
\]
\end{Remark}
Another idea is to quantify the obstruction to $T$ 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 $x \in R(H)$
\[
T(x)T(y) = T(xy)T(1)
\]
\end{Lemma}
This gives some hope for calculations when, for example, $H$ is a $p$-Sylow: in that case $T(1) = \CC\left[G/H\right]$ is $G/H$-torsion in the graded ring $R^*(G/H)$ and would be "somewhat invertible" in $R^*(G)$. Note also that $y$ being stable by $G$ seems a nontrivial condition: it is not satisfied by any irreducible representation of $H=C_p$ in the example above.
\begin{proof}
If $y$ is stable by $G$ then the induced character $T(y)$ is $T(1)y'$, 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:
\[
T(x)T(y) = T(x)T(1)y' = T(x \res(y'))T(1) = T(xy)T(1).
\]
\end{proof}
