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

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

  2. $AB = BC = CA = G$;

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

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

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A tiny remark is that for finite $G$, this can't happen with $A,B,C$ all nilpotent. If all were nilpotent, then $G$ is solvable by Kegel-Wielandt, and then $G$ has a unique conjugacy class of self-normalizing nilpotent subgroups (the Carter subgroups). Thus $A,B,C$ are all conjugate. But a group is never a product of two proper conjugate subgroups. – Geoff Robinson Jun 24 '11 at 15:37
Another comment in the finite case: Obviously we have $|A|=|B|=|C|$, and so $|G|$ is a perfect square, and each of $A,B,C$ is core-free. Even just the existence of an $A$ and a $B$ is equivalent to: $G$ is a transitive permutation group of degree $n$, order $n^2$, and has a self-normalizing regular subgroup. I don't know a lot about permutation groups, but I suspect this limits the structure quite a bit. If we could somehow show $|A^b\cap A|$ is independent of $b\in B$, we could even conclude $G$ is primitive. – Steve D Jun 24 '11 at 18:18
Just a minor remark: For searching the literature, it might be helpful to know that that when $AB=G$, $A\cap B=1$, then $G$ is also called an Zappa–Szép product, knit product or bicrossed product of $A$ and $B$. The group $G$ then also is referred to as a permutable group, or "group that admits an exact factorization". – Max Horn Jun 24 '11 at 20:55
up vote 16 down vote accepted

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 $ \left( \begin{array}{clcrc} 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 \end{array}\right)$. 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 $C_{U}(a) = I$. 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).

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I think the idea is good. But for $G\times G$ to be generated by $\Delta(G)$ and $\Delta^\alpha(G)$ you need that elements of the form $g^{-1}\alpha(g)$ generate $G$. Is it always true? There are many papers (after John Thompson) about groups admitting fixed point free automorphisms. – Mark Sapir Jun 25 '11 at 8:44
These two subgroups each have order $|G|$, so their product is $G \times G$ by order considerations, as I said. If a finite group $X$ has subgroups $A$ and $B$, then the set $AB$ has cardinality $|A| |B|/ |A \cap B|$, but need not be a subgroup. In this case $\Delta(G)$ and $\Delta^{\alpha}(G)$ each have order $|G|$ and have trivial intersection. The set $\Delta(G)\Delta^{\alpha}(G)$ has cardinality $|G|^{2}$, so must be all of $G \times G$. Similarly for other pairs. I know the literature on fpf automorphisms reasonably well. I think these triples should exist. – Geoff Robinson Jun 25 '11 at 9:13
Geoff: You are right. – Mark Sapir Jun 25 '11 at 9:33
@jp: Thanks for the smaller example. Second: yes, I should have said in the final write-up that given a fpf automorphism $\alpha$ of $X$ order $p^2$, then we do have $p$ self-normalizing subgroups of $X \times X$, each of order $|X|$, with pairwise trivial intersection. They are the elements of $\{ \Delta_{i}(X): 0 \leq i \leq p-1 \}$, where $\Delta_i(X) = \{(x, x\alpha^{i}): x \in X\}$. The point is that $\alpha^{i-j}$ is fpf for $i \neq j$ in this range. Also, once $\Delta(X)$ is self-normalizing, `$\Delta^{\alpha}(X)$ is, as it is conjugate to $\Delta(X)$ in ${\rm Aut}(X \times X)$. – Geoff Robinson Jun 26 '11 at 9:57
Actually, I think an example could be made to work just using ${\rm GL}(3,7)$, with $X$ of order $7^{3}.4$ admitting a fixed-point free automorphism $\alpha$ of order $9$ such that $\alpha^{3}$ centralizes a Sylow $2$-subgroup $U$ (a Klein 4-group) of $X$. The matrix representing $\alpha$ would just be $([0,1,0],[0,0,1],[2,0,0])$, and $U$ would consist of all diagonal unimodular matrices of order $2$. Can't see now why I felt I need to go to $6$-dimensions. Similarly, for other odd primes $p$, ${\rm GL}(p,q)$ should be enough for prims $q \equiv 1$ (mod $p$). – Geoff Robinson Jun 26 '11 at 18:52

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.

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