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Can a finite group G be the union of self normalizing subgroups such that the intersection between any two of these subgroups is equal to the unit of the group G? I don't think so but I can't prove it. Thank you for your help.

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2 Answers 2

The anwer is No.

Let $H_1,\dots, H_n$ be a family of subgroups of $G$ whose union equals $G$ and whose pairwise intersection is trivial. In the literature we call such a family a partition of $G$ by subgroups.

The partitions of all finite groups are explicitly known - in the solvable case thanks to Baer, and in the general case by Suzuki. The possibilities for partitions are listed at Example 3.5.1 in Schmidt's book Subgroup lattices of groups. (Email me if you want an e-copy of this book.)

You can go through the list of families and see, explicitly, that each partition contains a subgroup that is non-self-normalizing. In particular the four main examples are when $G$ is a Frobenius group, or when $G$ is isomorphic to $\mathrm{PGL}_2(q)$, $\mathrm{PSL}_2(q)$ or $\mathrm{Sz}(q)$; there is one more family which is built using one of these four families - since these four families are partitioned using non-self-normalizing subgroups, he last family will be also.

Note that this result on partitions doesn't use the Classification of Finite Simple Groups. It still uses some heavy-duty maths though, so one might still hope to find a more direct argument - the presence of the Suzuki groups in the classification suggests to me that such an argument would need to deal with the case when $G$ is simple, and then move to the general situation.

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To expand a little on Nick Gill's answer: it was a common theme in some early classification-type theorems that a simple group might be a disjoint union of Hall subgroups with trivial intersection of any two of them. This occurs for example, in Suzuki's classification of simple CA-groups ( groups in which all non-identity elements have Abelian centralizers)- I do not know whether Suzuki was thinking about groups with a partition before he did that classification. The Suzuki groups are simple CN-groups ( nonidentity elements have nilpotent centralizers)-there are simple CN-groups with no such partition, eg ${\rm PSL}(2,7), {\rm PSL}(2,9), {\rm PSL}(2,17)$ - in these last groups, Sylow $2$-subgroups can have non-trivial intersections. Quite often, delicate character theory was needed in these classifications. Also, by Frobenius's theorem, if $G$ is a finite simple group, and $M$ is a nilpotent Hall subgroup of $G$ disojoint from its other conjugates, then $M$ can't be self normalizing (there are finite simple groups which have a Sylow $2$-subgroup which is a maximal subgroup, eg ${\rm PSL}(2,17)$). Hence it follows for general reasons(without CFSG but as Nick says, with some quite difficult group theory) that in a finite non-Abelian simple group which admits a partition by nilpotent Hall subgroups, these can't all be self-normalizing and disjoint from their other conjugates ( apart from the identity). However, at the moment I don't see how to tackle directly ( ie without CFSG, and with easier results than Nick quotes ) partitions of the type asked about here (even just for simple groups).

Later edit: In a completely different direction, (though this may be covered by Baer's work, which predates Carter subgroups), a finite solvable group $G$ is never a union of proper nilpotent self-normalizing subgroups : for $G$ has a unique conjugacy class of nilpotent self-normalizing subgroups, the Carter subgroups, and no finite group is the union of a single conjugacy class of proper subgroups (if $G$ itself is nilpotent, it has no proper self-normalizing subgroups at all).

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That's a cool fact about Carter subgroups. I didn't know about those. –  Nick Gill Sep 5 at 14:51

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