I encountered this result while reading a few things, and it was stated without reference. I am having a hard time finding a reference for it (or a simple proof), so maybe you can help me:


let $E$ be an elliptic curve over $\mathbb{Q}$, $\ell$ a prime number, and consider $\rho=\rho\_{E,\ell}$ the Galois representation of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ on the $\ell$-adic Tate module of $E$. Let $p\neq\ell$ be a finite prime number where $\rho$ is ramified. Let $\ell^n$ be the highest power of $\ell$ such that $\rho\mod\ell^n$ is unramified at $p$. 


Then $(\ell)^n=(c\_p)$ in $\mathbb{Z}\_{\ell}$, where $c\_p$ is the Tamagawa factor of $E$ at $p$ (the order of the component group of the special fiber at $p$ of the N\'eron model of $E$ over $\mathbb{Z}$).



Neron models and ramification
This result (and much more) is in Exposé IX *Modèles de Néron et monodromie* by A.Grothendieck (in SGA7), and more precisely in section 11.1. In particular what you want is exactly Proposition 11.2 thereof. 


The proof is not hard but it also far from being obvious (to me, anyway). Let me recall it briefly. A separate, quasi-finite scheme $X$ over $\mathbb Z\_{p}$ decomposes as a disjoint union of a finite part $X^{f}$ which has same special fiber as $X$ with a scheme $X'$ with zero special fiber. Denote by $\mathcal E$ the Néron model of $E$. The composition of $\mathcal E(\bar{\mathbb Q}\_{p})^{f}\rightarrow (\mathcal E\times\_{\mathbb Z\_{p}}\mathbb F\_{p})(\bar{\mathbb F}\_{p})\rightarrow \Phi$ is surjective and its kernel is $\mathcal E(\bar{\mathbb Q}\_{p})^{f,\circ}$. As multiplication by $\ell^n$ is surjective on $\mathcal E^{\circ}(\bar{\mathbb Q}\_{p})^{f}$, the Snake Lemma gives an isomorphism $\Phi[\ell^{n}]\simeq\mathcal E[\ell^n]^{f}/\mathcal E[\ell^n]^{f,\circ}$. The image of $(T\_\ell E)^f\rightarrow E[\ell^n]^f$ is $\ell$-divisible so in $E[\ell^n]^{f,\circ}$ as $\Phi$ is finite and coincides with this subspace by surjectivity of multiplication by $\ell^n$. Hence $(T\_\ell E)^{f}\otimes\mathbb Z/\ell^{n}\mathbb Z\simeq \mathcal E[\ell^n]^{f,\circ}$.


Finally, the finite part of $\mathcal E[\ell^n]$ is given by the invariants under the inertia group $I\_p$. Putting all this together, we get the following explicit description of the $\ell^n$-torsion of the component group and the result
$$\Phi[\ell^n]\simeq\frac{(T\_{\ell}E\otimes\mathbb Z/\ell^{n}\mathbb Z)^{I\_p}}{(T\_{\ell}E)^{I\_p}\otimes\mathbb Z/\ell^{n}\mathbb Z}.$$


