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Let $G$ be group and $G$ has a subgroup $H$ with infinite index.

Is it true that the number of subgroups of $G$ containing $H$ is infinite?

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3 Answers 3

up vote 5 down vote accepted

No. I'm sure there are far simpler examples, but the first that comes to mind is a Tarski monster. This is an infinite group $G$ in which every nontrivial proper subgroup has order $p$, for some fixed prime $p$. Any such subgroup thus has infinite index, but is not contained in any subgroup except itself and $G$.

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Tarski monster group is a very interesting. Can you give me some reference about subgroup of this group? –  Babak Miraftab Jan 29 '13 at 18:58
    
What's wrong with the references cited in the Wikipedia article I linked? –  Chris Eagle Jan 29 '13 at 19:02
    
You claimed that it is not contained in any subgroup except itself and $G$. I want to see more detail about this. –  Babak Miraftab Jan 29 '13 at 19:05
    
@Babgen: How many ways can one subgroup of order p contain another subgroup of order p? –  hardmath Jan 29 '13 at 19:21

I believe the following should provide a relatively simple example. Let $N$ be the additive group of the rationals, and $H$ the multiplicative group of the nonzero rationals. Let \begin{equation} G = N \rtimes H \end{equation} be the semidirect product, in which $H$ acts on $N$ by multiplication.

Now $H$ is clearly a subgroup of infinite index in $G$. But $H$ is maximal in $G$, that is, there is no subgroup $L$ such that $H < L < G$. In fact, if $H < L$, then $L$ contains a nonzero $a \in N = \mathbf{Q}$, so if $b \in N = \mathbf{Q}$ is nonzero, then applying the element of $H$ that multiplies by $b a^{-1}$ we see that $b = (b a^{-1}) \cdot a \in L$, so that $L$ contains $N$, and thus $L = G$.

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What's wrong with the rather widespread $\,\Bbb Q\,,\,\Bbb Q^*\,$ ...? –  DonAntonio Jan 29 '13 at 19:38
    
Nothing @DonAntonio, but on the one hand I wanted to have an $H$ as in the original question, and then I wanted to make clear whether I was taking an element in one group or in the other. –  Andreas Caranti Jan 29 '13 at 19:47

In general, your result is false, but if $H$ is a normal subgroup, then it is true. Indeed, there is a bijection between the subgroups of $G$ containing $H$ and the subgroups of $G/H$; but $G/H$ has finitely many subgroups iff $G/H$ is finite iff $H$ is of finite index in $G$.

In particular, the result is true for abelian groups.

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5  
Let me ask, since I'm not the answerer here: why the downvotes? Has the poster said something wrong, in which case a comment pointing out the mistake is called for? Or is the observation correct but so irrelevant to the question that some feel a need to discourage this behavior? Inquiring minds wonder... –  hardmath Jan 29 '13 at 18:57
3  
I think some rather bitter serial downvoter is amongst us...Some people rush to downvote whenevever is, or seems to them to be, wrong. –  DonAntonio Jan 29 '13 at 19:40

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