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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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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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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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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