<|endoftext|>ogeny $E\to E/\langle P\rangle$).
1. The graph consists of one vertex, the curve $E/{\mathbb{Q}}$ does not have any isogenous curves (other than itself), and $E({\mathbb{Q}})_{\text{tors}}$ is necessarily trivial as any torsion point $P\in E({\mathbb{Q}})$ would induce a rational isogeny.
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2. The case of two vertices, i.e., the case when $C(E) = C_{p}(E) = 2$ and $C_q(E)=1$ for all $q\neq p$, occurs for $p\in \{2, 3, 5, 7, 11, 13, 17, 19, 37, 43, 67, 163\}$:
E\_[1]{} \[r\] & E\_[2]{} = E\_1/P
Suppose first that $p>7$ and $C_p(E)=2$. If $q$ divides the order of the torsion subgroup of $E_1$ or $E_2$, then $q=p$. Now, Mazur’s theorem \[thm-mazur\] shows that $E_1({\mathbb{Q}})_{\text{tors}}=E_2({\mathbb{Q}})_{\text{tors}}$ must be trivial. Hence, the isogeny-torsion graph $L_2$ is of the form $([1],[1])$.
Now let $p$ be $3, 5,$ or $7$, suppose $C_{p}(E) = C(E) = 2$ and such that $\langle P\rangle\subseteq E(\overline{{\mathbb{Q}}})_\text{tors}$ is a ${\mathbb{Q}}$-rational subgroup of order $p$. Both $E_1$ and $E_2$ can have trivial torsion over ${\mathbb{Q}}$, simultaneously, giving $([1],[1])$ as its $L_2$ type. If $P\in E_1({\mathbb{Q}})_{\text{tors}}$, however, by Lemma \[lem-subsequent-rational-pts\], $E_2 = E / \langle P\rangle$ has no rational points of order $p$ and thus, $E_2({\mathbb{Q}})_{\text{tors}}$ is trivial. Thus, the $L_2$-type is $([p],[1])$.
Finally, if $C_{2}(E) = C(E) = 2$, then both isogenous curves have torsion subgroups isomorphic to ${\mathbb Z}/ 2 {\mathbb Z}$, by Lemma \[lem-orbit-of-rational-points\] and Lemma \[lem-2torspt-all-have-2torspt\].
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3. By our work in Section \[sec-whatgraphs\] (see also Cor. \[cor-uniquecyclicQrationalgroup\]), the case of $L_3$, that is, $C(E)=C_p(E)=3$ and $n(P)=p^2$, can only occur for $p=3$ or $5$. In such case, we have
E\_[1]{} \[r\] & E\_[2]{} = E\_1/P\[r\] & E\_[3]{}= E\_1/P
Since $E_2$ has two independent $p$-isogenies, for $p=3$ or $5$, it follows that the image of $\rho_{E,p}$ is contained in a split Cartan subgroup of ${\operatorname{GL}}(2,{\mathbb F}_p)$. If $E_2$ does not have rational $p$-torsion points, then neither $E_1$ or $E_3$ would have rational torsion points, by Lemma \[lem-subsequent-rational-pts\] (which would imply a rational $p$-torsion point on $E_2$), and the isogeny-torsion graph type is $([1],[1],[1])$. Otherwise, if $E_2$ has a rational $p$-torsion point, then Lemma \[p\^2 isogeny\] implies that either $E_1$ or $E_3$ have a rational point of $p$-torsion (and, clearly, a cyclic rational $p^2$-isogeny). Let us assume that $E_1$ has a rational $p$-torsion point. Since $E_1$ has a unique $p$-isogeny, it follows that $P_1=[p]P$ must be one of the $p$-torsion points defined over ${\mathbb{Q}}$. By Lemma \[lem-subsequent-rational-pts\], we