Automorphism group of elliptic curve in char 2 I'm trying to calculate the automorphism group of elliptic curve with $j$-invariant $0$ in a field $K$ of characteristic $2$.
Let $ Y^2Z+b_3YZ^2=X^3$ the elliptic curve.
The substitutions preserving this form are:
$$X=u^2X+s^2Z$$
$$Y=u^2sX+u^3Y+t$$
$$Z=Z.$$
Then, the automorphisms of $E$ have
$$u^3=1 \text{ with $u$ in $K^*$}$$
$$s(b_3+s^3)=0$$
$$s^6+tb_3+t^2=0.$$
So I have $24$ possible triplets $(u,s,t)$ forming a group with the composition law $$(u,s,t)*(v,\gamma,\delta) = (uv,u\gamma+\delta,u^2\gamma^2s+\delta+t).$$
Let
$$a=(\xi_3,0,0)\text{ has order 3}$$
$$-1=(1,0,b_3)\text{ has order 2}$$
$$i=(1,\sqrt[3]b_3,b_3\xi_3^2)$$
$$j=(1,\sqrt[3]b_3\xi_3,b_3\xi_3^2)$$
$$k=(1,\sqrt[3]b_3\xi_3^2,b_3\xi_3^2)$$
with $$i^2=j^2=k^2=ijk=-1.$$
So $Q_8$ and $Z/3Z$ are two subgroups.
How could I say that the group of automorphisms of $E$ is the semi-direct product of $Q_8$ and $\Bbb Z/3\Bbb Z$?
This is what I've been thinking:
let $\phi:\Bbb Z/3\Bbb Z\to Aut(Q_8)$  such that $\Bbb Z/3\Bbb Z$ acts on $Q_8$ with a permutation of $\pm i,\pm j, \pm k$ and fixing $\pm 1$
$$(\xi_3,0,0)*(1,s,t)*(\xi_3,0,0)^{-1}=(1,s\xi_3,t)$$
$$(\xi_3^2,0,0)*(1,s,t)*(\xi_3^2,0,0)^{-1}=(1,s\xi_3^2,t)$$
so I have $axa^{-1}=\phi(a)(x)$ for all $a\in \Bbb Z/3\Bbb Z \text{ and } x\in Q_8$.
Is this the presentation of the semi-direct product of $Q_8$ and $\Bbb Z/3\Bbb Z$?
I'm not sure it is enough to come to the conclusion.
 A: Your work in the post is enough to say that this group is a semi-direct product $Q_8 \rtimes \Bbb Z/3\Bbb Z$. By the definition of a semi-direct product (see Wikipedia, for example), all that we need to verify that a group $G$ is a semi-direct product of a normal subgroup $N$ and a subgroup $H$ is that $G=NH$ and $N\cap H=\{e\}$. In our case, $N=Q_8$ (you can check normality by the group structure you've written down in your post) and $H=\Bbb Z/3\Bbb Z$, verifying that $N\cap H=\{e\}$ is clear since $N$ consists of elements of order dividing $3$ and $H$ consists of elements of order dividing $4$, and $G=NH$ by noticing that after you left-multiply an arbitrary element of $G$ by an appropriate power of $a$, you get an element in $H$. We can also see that it's a nontrivial semidirect product (i.e., not $Q_8\times \Bbb Z/3\Bbb Z$) by observing that $H$ isn't also normal - you have enough information to check this based on the group structure you've written down. So $G$ is a semi-direct product of $Q_8$ with $\Bbb Z/3\Bbb Z$.
Now we might ask whether there are different nontrivial semi-direct products $Q_8 \rtimes \Bbb Z/3\Bbb Z$. For this, we note that if $\varphi:H\to Aut(N)$ and $f$ is an automorphism of $N$ so that conjugation by $f$ is an automorhpism of $Aut(N)$ denoted by $\gamma_f$, then $N\rtimes_\varphi H$ and $N\rtimes_{\gamma_f\circ\varphi} H$ are isomorphic as groups. In our case, $Aut(Q_8)=S_4$ (see groupprops, for instance), and the image of $\Bbb Z/3\Bbb Z$ can be given by the span of some 3-cycle in $S_4$. Since there's an inner automorphism of $S_4$ sending any 3-cycle to any other 3-cycle, we see that any two nontrivial semidirect products $Q_8 \rtimes \Bbb Z/3\Bbb Z$ are isomorphic, so it makes sense to speak of "the semidirect product" here, and indeed what you have written down is this. (The question of when two semi-direct products are isomorphic can be a little difficult in general.)
A: For the benefit of people who find this question in the future, I note that the automorphism group is a familiar group, namely it is isomorphic to $SL_2(3)$.  This follows from KReiser's answer, since that answer shows there is a unique nontrivial semidirect product $Q_8\rtimes C_3$, and it's easy to check that $SL_2(3)$ is such a semidirect product.
