Finite subgroup scheme and Neron model of an abelian variety Let $p$ be a prime. Let $K/\mathbb{Q}_p$ be a finite extension and $\mathcal{O}=\mathcal{O}_K$ its ring of integers.
Let $A_K$ be an abelian variety and $\mathcal{A}_\mathcal{O}$ denote its Neron model over $\mathcal{O}$. We don't assume that $\mathcal{A}_\mathcal{O}$ has good or semistable reduction.
Let $\ell$ be a prime different from $p$ and let $G=A[\ell]_\mathcal{O}$ be the scheme-theoretic kernel of multiplication by $\ell$ in $\mathcal{A}_\mathcal{O}$. We know that $G\times_{\text{Spec} ~\mathcal{O}} \text{Spec} ~K = A_K[\ell]$ is a finite subgroup scheme of order $\ell^{2g}$, where $g$ is the dimension of $A_K$. 
I have a very stupid question here:
Is $A[\ell]_{\mathcal{O}}$ finite and flat? (or at least quasi-finite)? 
If not, are there some necessary conditions for $A[\ell]_{\mathcal{O}}$ to be finite and flat?
 A: In general the scheme $\mathcal A[l]_{\mathcal O}$ is not finite because of the following lemma.

Let  $f:X\to Y$ be a separated quasi-finite flat morphism of noetherian schemes. Then it is finite iff the fibral rank is locally constant.

For a proof see the paper "Les schémas de modules des courbes elliptiques" by Deligne and Rappoport (II$.1.19$). 
So, you actually need just to compute the fibral rank of $\mathcal A[l]_{\mathcal O}$. For example, if $A_K$ is an elliptic curve with additive or multiplicative reduction it is usually not finite. Note that there is a small mistake in Jason Starr's comment, the rank of $\mathcal A[l]_k$ might be bigger than $l$ in the case of multiplicative reduction b/c $\mathcal A[l]_k$ isn't necessary connected.
However, $\mathcal A[l]_{\mathcal O}$ is always quasi-finite provided that $l$ is invertible on $\mathcal O_K$. Moreover, it is always étale and you don't need any semi-stability assumptions for this claim. For a proof look at Lemma 7.3/2 in the book "Néron Models" by Bosch, Lutkebohmert, Raynaud.
