TITLE: Do isogenies with rational kernels tend to be surjective?
QUESTION [18 upvotes]: Dear MO Community, 
this is a pretty vague title, so let me tell you the precise observation I have made.
Consider the family of elliptic curves over $\mathbf{Q}$ having a rational $5$-torsion point $P$. They are given by
$$E_d: Y^2 + (d+1)XY +dY=X^3+dX^2,$$
for $d \in \mathbf{Q}^*$ and $P=(0,0)$. Let $\eta: E_d \rightarrow E_d'$ the isogeny whose kernel consists exactly of the five rational $5$-torsion points. 
Now assume that $E_d$ has rank $1$. After modding out torsion the Mordell-Weil group is isomorphic to $\mathbf{Z}$, and hence, $\eta$ induces an injective group homomorphism $\mathbf{Z} \rightarrow \mathbf{Z}$ which is either an isomorphism or has cokernel of size $5$.

It seems to me that this map tends to be an isomorphism.

To be precise, among all $d$, such that the numerator and denominator is bounded by $100$, there are $3,038$ elliptic curves of analytic rank equal to $1$ (out of $6,087$ total curves), and among those the above map $\mathbf{Z} \rightarrow \mathbf{Z}$ is an isomorphism in $91.2\%$ of the cases. 

So I wonder, what you should expect on average? $50\%$? $100\%$?

Maybe, this was just a coincidence in a small database. Maybe someone has seen a similar behaviour somewhere else. Maybe this is nothing new and I just haven't heard about it. I am curious to read what you think about it.
Many thanks.

REPLY [2 votes]: I tried the same as Chris, but I don't think it gives you an argument for the surjectivity. As it is too long for a comment, I post an answer.
Assume we are given an elliptic curve $E$ over $\mathbf{Q}$ with rational $5$-torsion and denote by $\eta:E \rightarrow E'$ the isogeny modding out the rational $5$-torsion. As I mentioned in my question this curve is given by a rational non-zero number $d$.
Write $d=u/v$, with $u$ and $v$ coprime non-zero integers. Now we apply the equation of Cassels and Tate which encodes the invariance of BSD:
$$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = \frac{R_{E'}}{R_E} \cdot \frac{ \# E(\mathbf{Q})^2_{tor} }{\# E'(\mathbf{Q})^2_{tor} } \cdot \frac{P_{E'}}{P_E} \cdot \prod_{p \leq \infty}\frac{c_{E',p}}{c_{E,p}}.$$
Now assume, that the elliptic curve has rank 1.
As Chris already stated, the regulator quotient equals $5^a$, for $a \in \{\pm 1 \}$, where $a=1$ if and only if the induced map of $\eta$ on the free part of the Mordell-Weil group is surjective. Hence this is the case we are interested in.
For the torsion quotient we have, that it equals $5^b$, for $b \in \{0,2 \}$, where $b=2$ if and only if $d$ is not a fifth power. This is true for $100\%$ of the cases.
For the periods and Tamagawa numbers we have, that they are equal to $5^{c-1}$, for
$$c=\# \{p \equiv 1(5),\ p \mid u^2+11uv-v^2\} + \# \{p=5,\ p^3 \mid u^2+11uv-v^2 \}$$
$$-\# \{p,\ p\mid uv \},$$
where $p$ ranges over the finite primes. (The results on the torsion quotient and on the period and Tamagawa quotient (= local quotient) can be found here http://arxiv.org/abs/1206.1822/)
(The case $5^3\mid u^2+11uv-v^2$ happens if and only if $u \equiv 7v (25)$.)
Hence, in 100% of the cases we have 
$$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = 5 \cdot 5^{\{\pm 1 \}} \cdot 5^c.$$
I don't see any reason why a negative value of $c$, which you should expect quite often, should force $a$ to be $1$, and not be (completely) subsumed in the Sha quotient.
So, instead of having a pro argument for $a=1$, this is just a pro argument, that the isogeny $\eta$ can alter the $5$-primary part of Sha in an arbitrary way.
(In case you wonder whether $c$ can be positive, look at $u=1$, $v=76971487$. Then $uv$ is prime and $u^2+11uv-v^2$ factors as $-1 \cdot 11 \cdot 31 \cdot 41 \cdot 61 \cdot 131 \cdot 151 \cdot 331 \cdot 1061$, hence $c=7$.)
(In my database of 3,038 curves, there is the following situation: $c=0$ for 247 curves. For all of them $\eta$ is NOT surjective. $c=-2$ for 2258 curves. For 20 of them $\eta$ is NOT surjective. $c=-4$ for 533 curves. For all of them $\eta$ is surjective.)