I have a question about Hecke operators which should not be too difficult. I'm reading these notes by Jerry Shurman on translating modular forms to the adeles and don't understand a certain sentence.
We fix an integer $L \geq 1$. For $p$ prime, we set
$$\Delta_p(L) = \{ \delta \in \operatorname{Mat}_2(\mathbb Z) : \operatorname{det}(\delta) = p, \delta \equiv \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \pmod{L\mathbb Z}\}$$
For $v$ a place of $\mathbb Q$ corresponding to a prime $p$, we let
$$K_{v,L} = \{ g \in \operatorname{GL}_2(\mathbb Z_p) : g \equiv 1 \pmod{L\mathbb Z_p} \}$$
$$K_{f,L} = \prod\limits_{v < \infty} K_{v,L}$$
Now fix a prime $p$ corresponding to a place $v$, and let $\delta \in \Delta_p(L)$. Let $\delta_f$ be its diagonal image in $\prod\limits_{v < \infty}' \operatorname{GL}_2(\mathbb Q_p)$.
In the notes it's claimed that $\delta_w^{-1} \in K_{w,L}$ for all places $w \neq v$. I'm missing something obvious because I don't see why this should be. If we let $\delta = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then
$$\delta^{-1} = \begin{pmatrix} dp^{-1} & -bp^{-1} \\ -cp^{-1} & ap^{-1} \end{pmatrix}$$
If $q$ is the prime corresponding to $w$, then we would need to show
$$ap^{-1} \equiv dp^{-1} \equiv 1 \pmod{L\mathbb Z_q} \tag{1}$$
$$bp^{-1} \equiv cp^{-1} \equiv 0 \pmod{L\mathbb Z_q} \tag{2}$$
(2) is immediate, since $L$ divides $b$ and $c$ in $\mathbb Z$. Since $d \equiv p \pmod{L\mathbb Z}$, we do have $dp^{-1} \equiv 1 \pmod{L\mathbb Z_q}$, as $p$ is a unit in $\mathbb Z_q$. The issue is $ap^{-1} \equiv 1 \pmod{L\mathbb Z_q}$. I don't see why this should be.
