# There exists an infinite group with only one subgroup of infinite index.

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?

• I think there are such groups, certain subgroups of $\mathbb{Q}/\mathbb{Z}$ – Orest Bucicovschi Feb 17 '15 at 8:46
• $\mathbb Z$ ?!? – MooS Feb 17 '15 at 9:32

## 2 Answers

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 $A = \mathbb Z$.

• It's false that it's the only such abelian group. You can take $(\mathbb{Q},+)$ – Tryss Feb 17 '15 at 10:50
• I can name you lots of subgroups of infinite index in $\mathbb Q$. – MooS Feb 17 '15 at 12:42
• Thanks for the explanation. – St Vincent Feb 17 '15 at 23:32

No, the above proposition is true. How about the direct product of $(\mathbb{Z},+)$ with the group $(\{1,-1\},\times)$.

• What is the unique subgroup $H$ of $(\mathbb{Z},+) \times (\{1,-1\}, \times)$ such that $|H| = n$ for some $n \in \mathbb{N}$ and $[(\mathbb{Z},+) \times (\{1,-1\}, \times):H] = \infty$? – St Vincent Feb 17 '15 at 8:42
• $H=\{(0,1),(0,-1)\}$ is a subgroup – Tryss Feb 17 '15 at 8:45
• You mean 'the' subgroup instead of 'a' subgroup, right? – St Vincent Feb 17 '15 at 8:47
• 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 : $(a,b)*(c,d) = (a+b,c\times d)$ – Tryss Feb 17 '15 at 8:52
• I see, so the operations are defined component-wise. – St Vincent Feb 17 '15 at 8:58