\subsection{Torsion Subgroup of $J_1(29)$}

Let $J=J_1(29)$. The paper \cite{conrad-edixhoven-stein:j1p} proves that 
$$
  2^6\cdot 3\cdot 7\cdot 43 \cdot 17837 \mid \#J(\QQ)_{\tor} 
    \mid 2^{12}\cdot 3\cdot 7\cdot 43 \cdot 17837
$$
(see the remark at the end of \S6.2.3 and Table~1). 
Also \cite[Conj.~6.2.2]{conrad-edixhoven-stein:j1p} conjectures that
in fact
$$\#J(\QQ)_{\tor} = 2^6\cdot 3\cdot 7\cdot 43 \cdot 17837,$$
though this is not proved in \cite{conrad-edixhoven-stein:j1p}.

\begin{proposition}
We have 
$\#J(\QQ)_{\tor} = 2^{???}\cdot 3\cdot 7\cdot 43 \cdot 17837.$
\end{proposition}
\begin{proof}
  It obviously suffices to prove that $\#J(\QQ)(2) = 2^{???}.$ From
  \cite[Table~4]{conrad-edixhoven-stein:j1p} we see that there are
  $\Q$-simple abelian subvarieties $A_1$, $A_2$, $A_3$, and $A_4$ of
  $J$ and an isogeny $\varphi: A_1 \times A_2 \times A_3\times A_4 \to
  J,$ with $A_3$ of dimension $6$ and $A_4$ of dimension $12$.
  Moreover, $\#A_3(\Q)(2) \mid 2^6$ and $\# A_4(\Q)(2)\mid 2^6$, where
  we compute the torsion multiple $T$ from columns 4 and 5 of the
  table.

  The 2-group $J(\QQ)(2)$ maps to 0 in the quotient $J/(A_3 + A_4)$,
  since that quotient is isogegnous to $A_1 \times A_2$, and the
  torsion multiple for $A_1\times A_2$ is odd (and the torsion
  multiple is an isogeny invariant).  Thus 
$J(\QQ)(2) = (A_3 + A_4)(\QQ)(2)$. 
Let 
$$B = (A_3 \times A_4)/(A_3\cap A_4) \isom A_3 + A_4 \subset J.$$
Consider the exact sequence
$$
 0 \to A_3 \cap A_4 \to A_3 \times A_4 \to B \to 0,
$$
and take Galois cohomology to obtain the exact sequence
$$
  0 \to A_3(\QQ)(2) \cap A_4(\QQ)(2) \to A_3(\QQ)(2) \times A_4(\QQ)(2) \to B(\QQ)(2)
   \to \H^1(\QQ, A_3 \cap A_4)(2) \to \H^1(\QQ, A_3\cross A_4)(2) \to \cdots
$$

By a computation using modular symbols, we find that
$$
  A_3 \cap A_4 = ...
$$
I don't know yet..., but at least it has order $2^8$.
Using modular symbols I think I can compute $A_3(\QQ)(2)$
and $A_4(\QQ)(2)$ on the nose, as long as they both turn out
to have order $2^6$, which is what I suspect. 
If they are both $\F_2^6$ (my guess), then


We have an sequence of maps 
$$ 
  0 \to A_3 \times A_4 \ra B \isom B^{\vee} \to A_3^{'} \times A_4' \to 0,
$$



which induces a (not exact) sequence of maps
$$ 
0 \to A_4(\QQ)(2) \to B(\Q)(2) \to A_3^{'}(\QQ)(2) \to \H^1(\QQ,A_4)(2) \to \H^1(\QQ,B)(2) \to \cdots 
$$

Using modular symbols, {\em I bet (!)} I can show the induced map
$$
  A_4(\QQ)(2) \to A_3^{'}(\QQ)(2)
$$
is an isomorphism (and likewise with $3$ and $4$ swapped).  
That would imply that $B(\Q)(2) \to A_3^{'}(\QQ)(2)$ is surjective, so the
map 
$$
  \H^1(\QQ,A_4) \to \H^1(\QQ,B)
$$ is injective.



\end{proof}
