As the title says, why is $$\wedge ^2 E[p] \cong \mu_p,$$ where $E[p]$ refers to the $p$-torsion points of an Elliptic curve over a number field $K$, $\wedge$ refers to the exterior product and $\mu_p$ refers to the group of $p^{th}$ roots of unity.
I know that for the formal multiplicative group $\mathbb{G}_m$ over $K$, $\operatorname{Tor}\mathbb{G}_m(K)$ contains all the roots of unity in $K$. I know you can also define formal group laws on elliptic curves. Was wondering if that is the direction to go?
Let $k$ be of characteristic coprime to $p$ and $E/k$.
The first thing to note is that $E[p]$ is an $\mathbb{F}_p$-vector space of rank $2$. The exterior square of $E[p]$ is therefore an $\mathbb{F}_p$-vector space of rank $1$ (spanned by the wedge of the independent basis vectors). Indeed, this is the case for any $k$-vector space of dimension $2$.
The non-trivial thing here is that the Galois action is really making $\bigwedge^2 E[p] \cong \mu_p$ as group schemes (or if you prefer as $\mathbb{F}_p$-vector spaces with a $\operatorname{Gal}(\bar k/k)$ action).
Recall that elements of $\bigwedge^2 V$ correspond to alternating bilinear forms on $V$ (were $V/K$ is a $K$-vector space). It comes equipped with a natural Galois action.
The Weil pairing is an alternating, symmetric, bi-linear, non-degenerate form $e_p : E[p] \times E[p] \to \mu_p$ and it is Galois equivariant. You can see this because the Weil pairing is a non-trivial alternating Galois equivariant bilinear form, and it therefore spans $\bigwedge^2 E[p]$ (it is a non-trivial element of a rank $1$ vector space). The Galois action follows by construction.