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.
Automorphism group of elliptic curve in char 2
algebraic-geometryautomorphism-groupelliptic-curvesgroup-theorysemidirect-product
Related Solutions
Edit: 16 vertices is indeed the smallest possible size for such a graph (see the proof below).
Here's an example of a graph with 16 vertices whose automorphism group is $Q_8$. As we will show, the automorphisms of the graph preserve the colors and arrows:
The symmetry of this graph is similar to the symmetry of the Cayley graph of $Q_8$ shown in the question above. Here's an argument that it's possible to recover the colors and arrows from the structure of the graph itself:
Cyan vertices have degree 7, while yellow vertices have degree 5.
Black edges connect pairs of yellow vertices.
Green edges connect vertices of different colors and do not share triangles with black edges.
Purple edges connect cyan vertices and are involved in triangles with green edges.
Red edges connect vertices of different colors and are involved in triangles with green and purple edges.
Blue edges connect vertices of different colors and are neither red nor green.
Cyan edges connect cyan vertices and are not purple.
Each black edge is involved in exactly one black-red-blue triangle, and the arrow points from the vertex opposite the red edge to the vertex opposite the blue edge.
Using the colors and arrows, it's easy to prove that any automorphism of this graph that fixes a vertex must fix all 16 vertices, so there are at most eight automorphisms. It's easy to check that there are in fact eight automorphisms, which form a copy of $Q_8$.
Incidentally, Babai constructs a 16-vertex graph for $Q_8$ in his paper "On the minimum order of graphs with given group" with 52 edges. The graph above is a slight improvement on this, having the same number of vertices but only 48 edges. Though 16 is the minimum possible number of vertices, it would be interesting to know whether there's a 16-vertex graph with fewer than 48 edges that has symmetry group $Q_8$.
Edit: 16 vertices is indeed the minimum possible. Indeed, any graph whose automorphism group is $Q_8$ must have at least two orbits of size $8$.
To see this, observe first that any faithful action of $Q_8$ on a graph must have at least one orbit of size $8$, since otherwise the action factors through $Q/\{\pm 1\}$.
Now, suppose we have a faithful action of $Q_8$ on a graph $G$, and suppose that $G$ has just one orbit of size $8$. Let $v$ be a vertex in this orbit. I claim that switching $v$ with $-1v$ is an automorphism of $G$. This automorphism is not in $Q_8$, since the action of $Q_8$ on the orbit of size $8$ is the left regular action, so it follows that the automorphism group of $G$ is larger than $Q_8$.
To prove that switching $v$ and $-1v$ is an automorphism of $G$, observe that:
If $w$ is any vertex not in the orbit of size $8$, then the stabilizer of $w$ is a proper subgroup of $Q_8$, so $-1w = w$. Thus, there is an edge from $v$ to $w$ if and only if there is an edge from $-1v$ to $w$.
If there is an edge from $v$ to $iv$, then its image under $i$ is an edge from $iv$ to $-1v$. Thus $v$ is connected by an edge to $iv$ if and only if $-1v$ is connected by an edge to $iv$. The same holds for $-iv$, $\pm j v$, and $\pm k v$.
This proves the claim, and therefore 16 is the smallest possible size.
After a while of thinking, I figured this out:
$f$ is unramified, so $f^{-1}(P_0)$ is two points. Because $f$ is also a group homomorphism, we must have that one of those points is $P_0$, and the other point is $P_1$, a 2-torsion point. Letting $P_2$ and $P_3$ be the other 2-torsion points, we must have that $f(P_2)=f(P_3)$, since $P_2+P_1=P_3$, so we can arrange $\pi$ to map $P_0$ to $\infty$ and $f(P_2)$ to $0$. Therefore the composite $\pi\circ f$ is given by a rational function with double poles at $P_0$ and $P_1$ and double zeroes at $P_2$ and $P_3$. If we make $g$ the map $x\mapsto x^2$, this means we need to find a rational function on $X$ with simple poles at $P_0$ and $P_1$ and simple zeroes at $P_2$ and $P_3$. If we embed $X\subset\Bbb P^2$ with equation $y^2z=x(x-z)(x-\lambda z)$ so that $P_1$ is $[0:0:1]$, then the rational function $y/x$ works, which you can see by checking the intersections $V(x)\cap X$ and $V(y)\cap X$.
Since any rational function is determined up to a constant by it's zeroes and poles counted with multiplicity, this shows that some multiple of $y/x$ is the required rational function.
Best Answer
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.)