Assume that an elliptic curve $E$ over $\Bbb Q$ has a reducible mod $p$ representation. Then

## Q: Why is the semi-simplification of $E[p]$ the direct sum of ${\Bbb Z}/p{\Bbb Z}$ and $\mu_p$?

Next

## Q: How can we find a curve $E'$ isogenous to $E$ such that $E'$ has a point of order $p$ as well as two independent rational point of order $2$.

This occurs in checking that for prime $p > 5$, the mod $p$ representation $$ \rho_{E_{abc},p} \colon {\mathrm{Gal}}(\overline{{\Bbb Q}}/{\Bbb Q}) \to {\mathrm{GL}}_2({\Bbb F}_p) $$ is irreducible, where $E_{abc}$ is Frey-Hellegouarch curve $E_{abc} \colon y^2 = x(x - a^p)(x + b^p)$ associated to triple $a,b,c \in {\Bbb Z}$ such that $a \equiv 3 (4), b \equiv 0 (2)$, and $abc \not=0$.

To show this, assume to the contrary that $\rho_{E_{abc},p}$ is reducible, then the semi-simplification of $E_{abc}[p]$ should be the direct sum of ${\Bbb Z}/p{\Bbb Z}$ and $\mu_p$. Then, we have $E'$ isogeneous to $E_{abc}$ having the property that $E'$ has a rational point of order $p$ as well as two independent rational point of order $2$, which will be shattered by Mazur's theorem. Thus we see $\rho_{E_{abc},p}$ is irreducible as desired.

I cannot understand this reasoning, which caused my questions. Pierre