I have already asked this question on stackexchange without much luck. I apologize if the question is too trivial to be asked here.

Let $A$ be an abelian variety defined over a number field $K$, $P \in A(K)$ a point of infinite order. Let $K_{\ell^n} = K(A[\ell^n])$ be the field of definition of the points in $A[\ell^n]$ and $K_{\ell^n, P} = K_{\ell^n}(Q)$ be the field obtained by extending $K_{\ell^n}$ such that the point $Q \in A(\overline{K})$ satisfying $\ell^nQ = P$ is defined over $K_{\ell^n, P}$.

Let $\nu$ be a prime of $K$ that splits completely in $K_{\ell^n}$ but not in $K_{\ell^n, P}$. Moreover assume that $A$ has good reduction at the prime $\nu$. Let $k_{\nu}$ be the residue field of $K$ at the prime $\nu$ and $A_{\nu}$ be the reduction of the abelian variety $A$ at the prime $\nu$. Then there is a reduction homomorphism $$ A(K) \xrightarrow{sp_{\nu}} A_{\nu}(k_{\nu}) $$ I have read without proof the following statement :

order of $sp_{\nu}(P)$ is divisible by $\ell$.

Any reference containing a proof or complete proofs are much appreciated. Thanks!

Update

Using the hint in the comment by Prof. Felipe Voloch, I have tried to outline an argument.

As suggested in the comment, we assume that order of $sp_{\nu}$ is not divisible by $\ell$ to arrive at a contradiction. Suppose $y \in A_{\nu}(k_{\nu})$, s.t. $\ell y = sp_{\nu}(P)$. Let $w$ be a place of $K_{\ell^n}$ above $v$ and $w'$ a place of $K_{\ell^n, P}$ above $w$. Let $Frob_{w}$ be the Frobenius in the Galois group $\mathrm{Gal}(K_{\ell^n, P}/K_{\ell^n})$. Then $y - sp_{w'}(Q) \in A_{\nu}[\ell] \subset A_{\nu}[k_{w}]$. This will imply that $sp_{w'}(Q) \in A_{\nu}(k_{w})$ and hence $Frob_w(sp_{w'}(Q)) - sp_{w'}(Q) = 0$.

Since $\ell Q = P \in A(K)$ and $Frob_w(P) = P$, we get that $Frob_{w}(Q) - Q \in A[\ell]$. Now, we use the fact that $A$ has good reduction at $v$ and hence also at $w$ because $K_{\ell^n}/K$ is unramified at $\nu$, to get that the $sp_{w}|_{A[\ell]}$ is an injective map. We already had $$sp_{w}(Frob_w(Q)-Q) = Frob_{w}(sp_{w'}(Q)) - sp_{w'}(Q) = 0.$$

Thus we get that $Frob_w(Q) - Q = 0$. If $w$ does not split entirely then that would mean that the action of $Frob_w$ should be non-trivial on $Q$. We arrive at a contradiction.