% \Chapter{Planar nearrings} % We recall the definition of planar nearrings and basic results (see \cite{Clay:Nearrings}). Let $(N,+,\cdot)$ be a left nearring. For $a,b \in N$ we define $a \equiv b$ iff $a\cdot n = b\cdot n$ for all $n\in N$. If $a \equiv b$, then $a$ and $b$ are called *equivalent multipliers*. A nearring $N$ is called *planar* if $| N/_{\equiv} | \ge 3$ and if for any two non-equivalent multipliers $a$ and $b$ in $N$, for any $c\in N$, the equation $a\cdot x = b\cdot x + c$ has a unique solution. A *Ferrero pair* is a pair of finite groups $(N,\Phi)$ such that $\Phi$ is a fixed-point-free automorphism group of $(N,+)$. Starting with a Ferrero pair $(N,\Phi)$ we can construct a planar nearring in the following way: Select representatives, say $e_{1},\ldots,e_{t}$, for some or all of the non-trivial orbits of $N$ under $\Phi$. Let $C = \Phi(e_1)\cup\ldots\cup\Phi(e_t)$. For each $x\in N$ we define $a\cdot x = 0$ for $a\in N\setminus C$, and $a\cdot x=\phi_{a}(x)$ for $a\in\Phi(e_{i})\subset C$ and $\phi_{a}(e_{i})=a$. Then $(N,+,\cdot)$ is a (left) planar nearring with $|N/_{\equiv}| = |\Phi|+1$. Every finite planar nearring can be constructed from some Ferrero pair together with a set of orbit representatives in this way. *The problem:* Find a planar nearring with 25 elements and 9 pairwise non-equivalent multipliers. *The solution:* We follow the Ferrero method described above for defining a nearring multiplication on an additive group. First we have to find a fixed-point-free (fpf) automorphism group of order $8$ on a group of order $25$. %and choose some orbit representatives. We start with the cyclic group of order $25$: First of all we ask for the existence of an fpf automorphism group on `CyclicGroup(25)' by computing an upper bound for its order. \beginexample gap> FpfAutomorphismGroupsMaxSize( CyclicGroup(25) ); [ 4, 1 ] \endexample This function returns a list with two integers, $4$ and $1$. The first number is an upper bound for the size of an fpf automorphism group; if there is a metacyclic fpf automorphism group, then it has a cyclic normal subgroup of index dividing the second number. These bounds are not sharp. % If the upper bound for the size of an fpf automorphism group on some group is $1$, we know that there is no nontrivial fpf automorphism group, no Ferrero pair, and no planar nearring on this group at all. Here, SONATA does not exclude the possibility that the cyclic group of order $25$ has an fpf automorphism group of order $4$. However, we can be sure that all fpf automorphism groups are cyclic and that none of them has size $8$. Thus we have to consider the elementary abelian group of order 25 instead. \beginexample gap> FpfAutomorphismGroupsMaxSize( ElementaryAbelianGroup(25) ); [ 24, 2 ] \endexample There might even exist an fpf automorphism group of order $24$. (In fact there is more than one. The reference manual explains how to obtain all nearfields of size $25$.) For our example, we could compute either a cyclic automorphism group or one isomorphic to the quaternion group with 8 elements. Let's try the latter. \beginexample gap> aux := FpfAutomorphismGroupsMetacyclic( [5,5], 4, -1 ); [ [ [ [ f1, f2 ] -> [ f1^2, f2^3 ], [ f1, f2 ] -> [ f2^4, f1 ] ] ], <pc group of size 25 with 2 generators> ] \endexample Here, the function `FpfAutomorphismGroupsMetacyclic' determines the metacyclic fpf automorphism groups on `AbelianGroup([5,5])' with generators $p,q$ satisfying $p^4 = 1, p^q = p^{-1}$, and $q^2 = p^2$. For each conjugacy class of such groups one representative is given. Conjugacy is determined within the whole automorphism group of `AbelianGroup([5,5])'. The actual output of the function is a list with 2 elements. The first is not the list of fpf groups up to conjugacy but the list of automorphisms $p,q$ generating those groups. The second element is simply the group `AbelianGroup([5,5])', on which the automorphisms act. Since there is only one pair of generators $p,q$, all fpf automorphism groups isomorphic to the quaternion group are conjugate. Now, we have our Ferrero pair $(G, \Phi)$. \beginexample gap> phi := Group( aux[1][1] ); <group with 2 generators> gap> G := aux[2]; <pc group of size 25 with 2 generators> \endexample %By picking any one non-zero element of $G$ as representative we can now %define a planar nearring according to the Ferrero method. Next we have to pick some orbit representatives. We note that for a fixed Ferrero pair distinct choices of representatives may yield isomorphic nearrings. The function `OrbitRepresentativesForPlanarNearRing' returns exactly one set of representatives of given cardinality for each isomorphism class of planar nearrings which can be generated from $(G, \Phi)$. \beginexample gap> OrbitRepresentativesForPlanarNearRing( G, phi, 1 ); [ [ f1 ] ] \endexample This tells us that all planar nearrings obtained from $(G,\Phi)$ with one orbit representative are in fact isomorphic. What happens if we choose $2$ representatives? \beginexample gap> reps := OrbitRepresentativesForPlanarNearRing( G, phi, 2 ); [ [ f1, f1*f2 ], [ f1, f1^2*f2^2 ] ] \endexample We obtain $2$ non-isomorphic planar near-rings. Let's just construct one of them. The result will be an `ExplicitMultiplicationNearRing'. \beginexample gap> n := PlanarNearRing( G, phi, reps[1] ); ExplicitMultiplicationNearRing ( <pc group of size 25 with 2 generators> , multiplication ) \endexample How many non-isomorphic planar nearrings can be defined from our Ferrero pair $(G,\Phi)$ in total? Since there are $3$ non-trivial orbits of $\Phi$ on $G$, we may choose up to $3$ representatives. \beginexample gap> Length(OrbitRepresentativesForPlanarNearRing( G, phi, 3 )); 6 \endexample Summing all up, we get exactly $9$ non-isomorphic planar nearrings with elementary abelian additive group of order $25$ whose multiplication is defined using a quaternion group of fpf automorphisms. %%% Local Variables: %%% mode: latex %%% TeX-master: "manual" %%% End: