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I was playing with the Hermitian curve $y^q + y = x^{q+1}$ over the field $GF(q^2)$ and chanced upon the following (non Abelian) group law on the points of the affine curve:

$(a,b) * (c,d) = (a+c,b+d+ac^q)$.

Over $GF(q^2)$ the group has $q^3$ points, the identity is $(0,0)$ and the inverse of $(a,b)$ is $(-a,b^q)$.

So my question is whether this is known. I could not find a reference. If you haven't seen this before and if you're interested in seeing how I found this group law, just shoot me an email.

Thanks for your time. Best, Hiren

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Yes, that group law is known. It is a disguised form of a Sylow $p$-subgroup of the automorphism group of the Hermitian curve (namely $\text{PGU}_3(q^2)$), where $p$ is the characteristic of $\mathbf{F}_q$. This group is well understood, for instance it's an extraspecial $p$-group.

For any $\mathbf{F}_{q^2}$-rational point $(a,b)$ on the curve $y^q+y=x^{q+1}$, the map $$ (x,y) \mapsto (x+a, y+a^qx+b) $$ is an automorphism of the curve. It is known that the set of all such automorphisms forms a group $G$, and that this group $G$ acts faithfully and transitively on the set $S$ of $\mathbf{F}_{q^2}$-rational points on the curve. Moreover, $G$ has order $q^3$, and is a Sylow $p$-subgroup of the automorphism group of the curve.

For any faithful transitive action of a group $G$ on a set $S$, we can pick a point $P\in S$ and identify the group element $g\in G$ with the point $g(P)\in S$. Then we obtain a group structure on $S$ via $g(P)*h(P):=gh(P)$, where the identity element is $P$. Your group law on the set $S$ of $\mathbf{F}_{q^2}$-rational points is obtained from the above group $G$ by putting $P=(0,0)$.

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  • $\begingroup$ Thanks for your answer. I found the group law as follows: let t(a,b) be the tangent line through the point (a,b). Then t(a,b) = y-a^qx +b^q. Check that: t(a,b)^q + t(a,b) = (x_a)^{q+1} where x_a = x-a. This equation is just the same as that defining the Hermitian curve with t(a,b) replacing y and x_a replacing x. Now let T(c,d) be the tangent line (in terms of t(a,b) and x_a) to this curve at the point (c,d). Then T(c,d) = t(a,b) - c^qx_a + d^q and one checks that T(c,d) equals t(a+c,b+d+ca^q). So, from the points (a,b) and (c,d) we 'get' the point (a+c,b+d+ca^q). This gives a group. $\endgroup$
    – Hiren
    Commented Dec 14, 2013 at 7:59

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