How do you calculate the group scheme of E[p] for a an elliptic curve E in characteristic p? I know that the answer is $\mu_p \times \mathbb{Z}/p\mathbb{Z}$ if $E$ is ordinary, and $\alpha_p$ if $E$ is supersingular, where $\mu_p$ and $\alpha_p$ are the kernels of Frobenius  on $\mathbb{G}_m$ and $\mathbb{G}_a$ respectively. But why is it this true? 
Suppose $E'$ is a lift of $E$ to characteristic 0. Then $E'[p] = (\mathbb{Z}/p\mathbb{Z})^2$. If $E$ is ordinary then we have $E[p] =\mathbb{Z}/p\mathbb{Z}$, and one way to reconcile these two facts is to have $E[p] =  \mu_p \times \mathbb{Z}/p\mathbb{Z}$, since the group of closed points of $\mu_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ in characteristic 0, whereas in characteristic p, it is just a single (nonreduced) point. I'm not sure why this is the only possibility though--I think it has to do with the height of the formal group, but I just can't nut out the details. In Katz-Mazur "Arithmeticic moduli of elliptic curves" (proof of theorem 2.9.3) they say that "any p-divisible group over an algebraically closed field is the product of a p-divisible commutative formal Lie group and  finite number of copies of $\mathbb{Q}_p/\mathbb{Z}_p$," but I don't see why this is true. 
For the supersingular case, I'm even hazier. Is $\alpha_p$ the unique one-paramater formal group of height 2, and if so, how can you see this? For an affine scheme Spec(R), we have $\alpha_p(R) = \mathrm{Spec}(R[x]/(x^p))$. Is it true that in characteristic 0 we have (for example) $\alpha_p(\mathbb{C}_p) = (\mathbb{Z}/p\mathbb{Z})^2$?
Sorry if this is a mess, i'm really confused, and I haven't been able to find a sufficiently dumbed down explanation of this stuff anywhere.
 A: Dear Maxmoo,
Just to offer a slightly different perspective than that given by Kevin and Brian:
While their advice is certainly correct, when I was learning this I also found it very
helpful to make a couple of "bare hands" computations, as a kind of reality check.
For this, begin with an elliptic curve in char. $2$, in fact with two, of the form:
$$y^2 + y = x^3$$
and
$$y^2 + x y = x^3 + x $$
One of these is supersingular, the other ordinary.  (I won't tell you which here!)
Now try computing the $2$-torsion concretely, using lines passing through three points
and so on.
Remember that in the end you are looking for a degree $4$ equation (you may need to change
variables to see the point at infinity; this won't show up in the affine equations I've
given you).  By general theory, you know this equation won't be separable: non-reduced
group scheme structure will show up concretely as inseparability in this polynomial.
In one case (the s.s. case) it will be entirely inseparable; in the other (ordinary) case
it will have inseparability degree $2$ (so "half" inseparable, "half" separable).
Once you've done the case of char. $2$, you might want to try char. $3$ as well (since
computing the equation for the 3-torsion is also just about in reach by hand).
The reason I suggest this is that I remember, when I was learning this stuff, that all
these group schemes (especially the non-reduced ones) seemed fairly ephemeral, but after
I had made these kind of explicit computations, I had a much more concrete mental model
for what the general theory was talking about, which gave me a lot more confidence in
reading and making arguments about these kinds of things.
Best wishes,
Matt
