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Prove or disprove: There exists an infinite group with only one subgroup of infinite index.

By Lagrange's theorem, a group $G$ for some $H \leqslant G$ is partitioned into $[G:H]$ many subsets, each with cardinality $|H|$. If $|H|$ is finite, then the index of $H$ must be infinite by this theorem. If $|H| = \infty$, then we could have either a finite index or an infinite index.

Since $G$ contains infinitely many subgroups, I can't think of case where this would be true, since the above implies that this would be a group with a single finite subgroup. Is the above proposition false?

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    $\begingroup$ I think there are such groups, certain subgroups of $\mathbb{Q}/\mathbb{Z}$ $\endgroup$
    – orangeskid
    Feb 17, 2015 at 8:46
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    $\begingroup$ $\mathbb Z$ ?!? $\endgroup$
    – MooS
    Feb 17, 2015 at 9:32

2 Answers 2

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Obviously $\mathbb Z$ is such a group. The only subgroup with infinite index is $0\mathbb Z$.

Let us further prove, that $\mathbb Z$ is the only such abelian group. Let $A$ be an infinite abelian group (additive notion) with $(0)$ being the only subgroup of infinite index. In particular we have $|(a)| = \infty$ for any $0 \neq a \in A$, hence $A$ is torsion free.

Furthermore we have an exact sequence $0 \to \mathbb Z \to A \to A/(a) \to 0$ for any $a \neq 0$. Since $\mathbb Z$ and $A/(a)$ are finitely generated, we obtain that $A$ is finitely generated.

But finitely generated and torsion free implies free, thus $A = \mathbb Z^n$ for $n \geq 1$. For $n \geq 2$, $\mathbb Z^n$ has many subgroups of infinite index. We obtain $A = \mathbb Z$.

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  • $\begingroup$ It's false that it's the only such abelian group. You can take $(\mathbb{Q},+)$ $\endgroup$
    – Tryss
    Feb 17, 2015 at 10:50
  • $\begingroup$ I can name you lots of subgroups of infinite index in $\mathbb Q$. $\endgroup$
    – MooS
    Feb 17, 2015 at 12:42
  • $\begingroup$ Thanks for the explanation. $\endgroup$
    – St Vincent
    Feb 17, 2015 at 23:32
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No, the above proposition is true. How about the direct product of $(\mathbb{Z},+)$ with the group $(\{1,-1\},\times)$.

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  • $\begingroup$ What is the unique subgroup $H$ of $(\mathbb{Z},+) \times (\{1,-1\}, \times)$ such that $|H| = n$ for some $n \in \mathbb{N}$ and $[(\mathbb{Z},+) \times (\{1,-1\}, \times):H] = \infty$? $\endgroup$
    – St Vincent
    Feb 17, 2015 at 8:42
  • $\begingroup$ $H=\{(0,1),(0,-1)\}$ is a subgroup $\endgroup$
    – Tryss
    Feb 17, 2015 at 8:45
  • $\begingroup$ You mean 'the' subgroup instead of 'a' subgroup, right? $\endgroup$
    – St Vincent
    Feb 17, 2015 at 8:47
  • $\begingroup$ Oh, sorry, you're not familiar with the notion of direct product group, I should have explained more. It's simply the operation coordinate by coordinate : $(a,b)*(c,d) = (a+b,c\times d)$ $\endgroup$
    – Tryss
    Feb 17, 2015 at 8:52
  • $\begingroup$ I see, so the operations are defined component-wise. $\endgroup$
    – St Vincent
    Feb 17, 2015 at 8:58

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