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We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4?

In this paper http://www.jstor.org/stable/2695649 by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$.

I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?

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Your questions are in strange places. The first reads like a statement of a theorem by the student in class who is unsure; the last one confused me until I noticed you were missing an eroteme. –  Joshua Shane Liberman Sep 4 '10 at 13:14
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2 Answers

up vote 3 down vote accepted

One way to ensure this happens is to have every maximal subgroup be characteristic. To get every maximal subgroup normal, it is a good idea to check p-groups first. To make sure the maximal subgroups are characteristic, it makes sense to make sure they are simply not isomorphic. To make sure there are not too many maximal subgroups, it makes sense to take p=2 and choose a rank 2 group.

In fact the quasi-dihedral groups have this property. Their three maximal subgroups are cyclic, dihedral, and quaternion, so each must be fixed by any automorphism.

So a specific example is QD16, the Sylow 2-subgroup of GL(2,3).

Another small example is 4×S3. It has three subgroups of index 2, a cyclic, a dihedral, and a 4 acting on a 3 with kernel 2. Since these are pairwise non-isomorphic, they are characteristic too. It also just so happens (not surprisingly, by looking in the quotient 2×2) that every element is contained in one of these maximal subgroups.

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Thanks for the answer. I got some more information! Please look at it! –  anonymous Sep 4 '10 at 13:30
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Update: While searching on the internet i found this paper, "A remark on Hyperabelian groups" by G.Baumslag where there is this remark

Moreover so is any torsion abelian group of finite rank, since it is the union of finite characteristic subgroups. Would like to get a reference for this.

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Be careful when reading about infinite groups. Baumslag is not claiming that every torsion abelian group is the union of proper characteristic subgroups, only of finite characteristic subgroups. For instance the Klein four group is not the union of proper characteristic subgroups (the only such subgroup being the identity subgroup), but it is the union of its finite characteristic subgroups (the whole group). If the torsion abelian group is finite rank, then G[n] = { g in G : g^n = 1 } are finite, and of course it is the union of those characteristic subgroups. –  Jack Schmidt Sep 5 '10 at 1:14
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