Groups with triple system of self-normalizing subgroups - MathOverflow

Groups with triple system of self-normalizing subgroups

2011-06-24T08:41:24Z

Does there exist a group $G$ (finite or infinite) with three subgroups $A, B, C \leq G$ satisfying the following three conditions?

$A = N_G(A)$, $B = N_G(B)$, $C = N_G(C)$;

$AB = BC = CA = G$;

$A \cap B = B \cap C = C \cap A = 1$.

(This question turned up in a more specific setting, but a negative answer to the existence of such a group would settle our case.)

Answer by Andreas Thom for Groups with triple system of self-normalizing subgroups

2011-06-24T20:16:32Z

This is just an idea, not a solution. In

Marc Burger, Shahar Mozes, Lattices in product of trees, Inst. Hautes Etudes Sci. Publ. Math. 2000 (92) pp. 151-194

the authors study a class of groups G which is generated by two free groups $A= \langle a_1,\dots,a_n \rangle$ and $B = \langle b_1,\dots,b_n \rangle$ such that relations $a_i b_j = b_k a_l$ hold for a suitable set $\Sigma$ of quadruples $(i,j,k,l)$. (The set $\Sigma$ must satisfy some natural conditions.) This is made in such a way that $G= A B$ and $A \cap B= \lbrace e\rbrace$. Choosing $\Sigma$ well, Burger and Mozes manage to construct a finitely presented, torsionfree, simple groups. This is one of the main achievements of the paper. The group $G$ arises as a lattice in the automorphism group of a product of trees.

I could imagine (but have not checked) that a suitable choice of $\Sigma$ will also give that $A=N_G(A)$ and $B=N_G(B)$. One can now try to set $C:= \langle a_1b_1,\dots,a_n b_n\rangle$ (or something like this). Then, $AC=G$, $BC=G$, and $A \cap C=\lbrace e\rbrace$ and $B \cap C = \lbrace e\rbrace$ for suitable $\Sigma$. In order to get $C=N_G(C)$, one probably needs again additional properties on $\Sigma$. But maybe this can work as the group $G$ can be studied by its action on the product of trees.

Answer by Geoff Robinson for Groups with triple system of self-normalizing subgroups

2011-06-24T21:49:39Z

I revise earlier edits to give a coherent account of the construction which shows that such subgroups can exist.

The underlying idea of the strategy is as follows: Let $X$ be a non-trivial finite group with trivial center which admits an automorphism $\alpha$ fixing only the identity ((informally, and by a slight abuse, known as fixed-point free automorphism). Note that $\alpha$ can't have prime order, for otherwise $X$ would be nilpotent by Thompson's theorem, and hence would have non-trivial center. Then the direct product $X \times X$ is a product of two self normalizing subgroups of order $|X|$ which intersect trivially. One is $\Delta(X) = \{ (x,x): x \in X \}$ . The other is $\Delta^{\alpha}(X) = \{ (x,x\alpha): x \in X\}$ . That $\Delta(X)$ is self-normalizing is clear, since $Z(X) = 1$. For notice that if $(x,x)^{(a,b)} \in \Delta(X)$ for each $x \in X$, then $ab^{-1} \in Z(X)$. The argument for the other subgroup is similar. Now we seek such a group $X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. If such a group exists, we may set $\beta = \alpha^{-1}$. Then the group $G = X \times X$ has the desired three self-normalizing subgroups $\Delta(X)$ , $\Delta^{\alpha}(X)$ and $\Delta^{\beta}(X)$ . For notice that if $x \alpha = x \beta$, then $\alpha^{2}$ fixes $x$, so that $\alpha$ fixes $x$ as $\alpha$ has odd order. But then $x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $X \times X$ by order considerations.

There does exist a group $X$ of order $7^{6} .2^{4}$ which admits a fixed point free automorphism $\alpha$ of order $9$, and which has trivial center. Hence the construction above does work in this case, and $G = X \times X$ has three self-normalizing subgroups of the required form. We find a subgroup $Y$ of order $144$ of ${\rm GL}(6,7)$ which has an elementary Abelian normal subgroup $U$ of order $16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the eigenvalue $1$ in the given representation. We then take $X$ to be the semidirect product $VU$, where $V$ is a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given elementary Abelian subgroup of ${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(6,7)$, and to act on $V$ as the given matrix yields an action of $a$ on $VU$ as a fixed-point free automorphism of order $9$. In terms of matrices, $a$ is the matrix $ ( [0,1 , 0 ,0 , 0 , 0],[0 ,0, 1 , 0 , 0 , 0], [2 , 0 , 0 , 0, 0, 0], [0,0, 0 , 0 ,1 , 0],[0 , 0 , 0 , 0 , 0 , 1],[ 0 , 0 , 0 , 4, 0, 0])$ . The group $U$ is the set of diagonal matrices with diagonal entries $\pm 1$ such that the product of the first three diagonal entries is $1$ and the product of the last three diagonal entries is $1$. The subgroup $U$ is $a$-invariant, $a^3$ centralizes $U$, but $C_{U}(a) = I$ . (Could not get latex right for matrix). The group $VU\langle a \rangle$ is the semidirect product $VY$. A similar construction works for other odd primes $p$ by considering ${\rm GL}(2p,q)$, where $q$ is a prime congruent to $1$ (mod p).