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Suppose $G$ is a finite group and $N_1, N_2, \cdots, N_k$ are proper normal subgroups of $G$. The set $\{ N_1, \cdots, N_k\}$ is called a normal cover for $G$, if $G = \cup_{i=1}^kN_i$. I need to the main properties of finite groups with a normal covering. Anybody knows some references on this topic. There are several published papers on covering of groups by subgroups, but I like normal cover. Any comments will be highly appreciated.

Sincerely, F Moftakhar

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    $\begingroup$ What kind of properties are you interested in? $\endgroup$ – Ryan Budney Oct 27 '14 at 15:18
  • $\begingroup$ A classification theorem or something like this. $\endgroup$ – Fatemeh Moftakhar Oct 27 '14 at 15:20
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    $\begingroup$ I don't understand why people are voting to close this. It seems a perfectly reasonable question, if a little vague. $\endgroup$ – Nick Gill Oct 28 '14 at 15:56
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The following paper is relevant:

Finite coverings by normal subroups by Brodie, Chamberlain and Kapp, PAMS 1988.

The main focus is on coverings of infinite groups although their main theorem is still interesting in the finite case.

Theorem. A group has a nontrivial finite covering by normal subgroups if and only if it has a quotient isomorphic to an elementary abelian $p$-group of rank two for some prime $p$.

One of the corollaries to this theorem is also relevant for finite groups:

Corollary. Let $G=\bigcup\limits_{i=1}^n N_i$ where $N_1,\dots, N_n$ form an irredundant covering of $G$ by proper normal subgroups. Then $G/D$ with $D=\bigcap\limits_{i=1}^n N_i$ is finite and solvable.

This corollary effectively reduces the question of coverings by normal subgroups to the study of solvable groups. The paper can be viewed here. If you don't have access email me and I'll send it to you.

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  • $\begingroup$ There is a paper by Bhargava which gives a relatively easy proof of this theorem if I'm not mistaken. $\endgroup$ – Steve D Nov 20 '14 at 5:10
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I don't know references, but the following might be a starting point for an inductive approach. You might as well assume that each $N_{i}$ is maximal normal in $G.$ Now take a minimal normal subgroup $M$ of $G.$ If $M$ is contained in each $N_{i},$ then you can pass to $G/M$ which still has a normal covering. So you might as well suppose that there is some $i$ with $M \not \leq N_{i}.$ Then $G = M \times N_{i}$ for that $i.$

Later edit: In fact, this reduces the question to the case that $G$ is a direct product of simple groups, some of which may be Abelian, as $G$ embeds into $G/N_{1} \times \ldots \times G/N_{k}$ when $\bigcap_{i=1}^{k} N_{i} =1.$

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This is perhaps not quite on point, but I would like to mention a pretty little fact that is at least related to the question. If $G = \bigcup N_i$, where $N_i \triangleleft G$ and the $N_i$ are nontrivial and proper and have trivial pairwise intersections, then $G$ is an elementary abelian $p$-group for some prime $p$.

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You can find a complete classification here:

M. Chiodo, Finitely annihilated groups, Bull. Austral. Math. Soc. 90, No. 3, 404-417 (2014).

This also contains a generalisation to arbitrary coverings of groups by proper normal finite-index subgroups; such a group is said to be Finitely Annihilated (F-A).

See in particular corollary 5.5: A finite group is F-A if and only if it has non-cyclic abelianisation.

-Maurice

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  • $\begingroup$ This Corollary 5.5 follows immediately from the theorem of Brodie, Chamberlain and Kapp, PAMS 1988, see Nick Gill's answer: mathoverflow.net/a/185604/24165 . $\endgroup$ – Anton Klyachko Nov 20 '14 at 1:13
  • $\begingroup$ Yes, you're right, it does. However, the preceding result in that paper requires a little more work, though still uses the Brodie-Chamberlain-Kapp result. Proposition 5.4. Let $G$ be a finitely generated group with only finitely many distinct finite simple quotients. Then $G$ is F-A if and only if $G^{ab}$ is noncyclic. $\endgroup$ – MCC Nov 20 '14 at 9:01

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