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Does there exist a group $G$ that satisfies the following conditions:

  • Any proper subgroup of $G$ is contained in a maximal subgroup.
  • There is some $N\unlhd G$ such that $\frac{G}{N}$ is divisible.

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How about if $G$ is abelian?

My definition of divisibility is the same as what is used in this post: https://math.stackexchange.com/a/419101/138171

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Minimus Heximus
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1 Answers1

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There does not exist an abelian group with the prescribed properties.

Assume contrariwise that $G$ is such a group and that $N\lhd G$ gives rise to a divisible quotient group. The first assumption implies that $N$ is contained in a maximal subgroup $M$. If $G/M$ is infinite, then it has non-trivial subgroups, so correpondence principle tells us that there are subgroups properly between $M$ and $G$. So $G/M$ is finite, say $n=[G:M]$.

This implies that for all $x\in G$ we have $x^n=M$. So if $yN\in G/N$ is chosen in such a way that $y\notin M$, then $(zN)^n\neq yN$ for all $z\in G$, because $(zN)^n\in M/N$.

If we look for a non-abelian example, the same argument does not work, because $M$ is not necessarily normal. The argument only disallows those pairs $(G,N)$ such that $N$ is contained in a normal subgroup of a finite index.

Jyrki Lahtonen
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  • if $x\in G$, then $Mx^n=M$ so $x^n\in M$. – Minimus Heximus Oct 05 '14 at 20:06
  • There is some $y\in G\setminus M$. for any $z\in G$, if $Nz^n=Ny$ then $yz^{-n}\in N\subseteq M$ and so $My=Mz^n=M$ which implies $y\in M$. So $Nz^n\ne Ny$. – Minimus Heximus Oct 05 '14 at 20:47
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    So the equation $\theta ^n = yN$ has no solution for $\theta$ in $G\over N$ which is a contradiction because $G\over N$ is divisible by hypothesis. thanks. I could accept this answer if the group was assumed to be abelian. – Minimus Heximus Oct 05 '14 at 20:55
  • A subgroup of index $n$ contains a normal subgroup of index $\le n!$, so in such a group $G$ there has to be a maximal subgroup of infinite index. – Jyrki Lahtonen Oct 05 '14 at 21:10
  • Do you have a reference for this proposition? – Minimus Heximus Oct 05 '14 at 21:26
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    @user It is a standard result: act on the cosets of your subgroup by left multiplication, and take the kernel of this action. – user1729 Oct 05 '14 at 21:57