You mean that we use the fact that (5.2.2) is equal to $\langle \phi(x), \langle u(y), \rho(x-y) \rangle \rangle$, right?

And how do we deduce that this is equal to $\langle \rho(x) \otimes u(y), \phi(x+y) \rangle$ ?

@EricStucky We have that $|G|=p^km$, where $k\not\mid m$. That's why a $p$-Sylow group is a group of maximum order $p^k$, right?

Since $N_G(P)\leq G$, we have that $|N_G(P)|\mid |G|$. Therefore, the prime factorization of $N_G(P)$ has no more than $k$ factors of $p$.

We have that $P\leq N_G(P)$, so $|P|\mid |N_G(P)|$. Therefore, $N_G(P)=p^kn$, where $k\not\mid n$ and so $P$ is a $p$-Slyow subgroup also of $N_G(P)$, right?

$Z'C = B$

hi... Does anyone have a feeling if the following could plausibly be true? Consider a real valued positive definite matrix $M$. Is the following inequality true?

$$\sum_{k\in\mathbb{Z}^n} e^{-k^T M k} \leq \prod_{i=1}^n \left( \sum_{x \in \mathbb{Z}^n} e^{-\lambda_i x^2} \right) \ .$$

trying to see how we obtain the exact sequence
$0 \to \tilde{H^1}(D) \to H^1(Y) \to H^1(T) \to 0 $

sorry i mean
$0 \to \tilde{H^0}(D) \to H^1(Y) \to H^1(T) \to 0 $