# Is there a proper subgroup $H$ of $(\mathbb{R},+)$ such that $[\mathbb{R}:H]$ is finite?

I would like to know if there is a proper subgroup $H$ of $(\mathbb{R},+)$ such that $[\mathbb{R}:H]$ is finite. I know if I pick up $\mathbb{Z}$, then $[\mathbb{R}:\mathbb{Z}]$ is infinite.

-
There are such $H$ with countable index (assuming AC), even subrings, as seen in the MathOverflow question Are there countable index subrings of the reals?. –  Jonas Meyer Jan 29 '12 at 8:02

We claim that there is no subgroup of finite index in $\mathbb R$. Here is the proof:

Proof:

Suppose $\mathbb R$ has a proper subgroup of finite index. That is there exists a proper subgroup $H$ such that, $[\mathbb R:H]=n$.

Since, $\mathbb R$ is an abelian group, $H$ is a normal subgroup. So, the group of cosets of $H$ in $\mathbb R$ has the property that $(rH)^n=H$ for each $r\in \mathbb R,$ which implies that $nr \in H$ for each $r \in \mathbb R.$

Note that, this property is a consequence of the Lagrange's Theorem: The order of an element in a finite group divides the order of the group.

From this, we claim that, $\mathbb R \subset H$, which will contradict that $H$ is a proper Subgroup of $\mathbb R$.

For each $r \in \mathbb R$, $\dfrac{r}{n} \in \mathbb R$. Now, we have, $n \cdot \dfrac{r}{n}=r \in H$. This proves the claim and hence the contradiction.

Aside:

This in fact proves something more: As ymar points to in his comments: Any divisible group cannot have proper subgroups of finite index. Now what are these groups: Intuitively, those groups in which you can divide! Formally, these are groups in which each element is an $n^{th}$ mutiple of some element for each $n \in \mathbb N$.

-
Just a little comment for spohreis's attention. Note that Kannappan has proved that no divisible group has a subgroup of finite index. –  user23211 Jan 28 '12 at 14:49

This observation is implicit in the other answer.

Let $A$ be a domain which is not a field, and let $M$ be a finitely generated divisible $A$-module. Then $M=0$.

Assume by contradiction $M\neq0$, and let $N$ be a maximal proper submodule of $M$. Such exists because $M$ is nonzero and finitely generated. Then $S:=M/N$ is isomorphic to $A/\mathfrak m$, where $\mathfrak m$ is a maximal ideal of $A$. As $A$ is not a field, $\mathfrak m$ contains a nonzero element $a$, and we have:

$\bullet\$ $0\neq S=aS$, because $S$ is divisible and $a$ is nonzero,

$\bullet\$ $aS=0$, because $a$ is in $\mathfrak m$.

-
My knowledge of Algebra is limited to some group theory and Ring Theory I know is a 3 weeks material taught as a Prelude to Linear Algebra Course. +1 anyway for writing out some observation which has missed a lot of people's eyes. I hope to understand it in the near future! –  user21436 Jan 29 '12 at 15:10
Dear @Kannappan: Thanks! +1 for your answer too! If you think I can help you in any way, I'd be glad to do it. –  Pierre-Yves Gaillard Jan 29 '12 at 16:22
I'll take help understanding this by pinging you here! But not right away with an assignment due in a few hours and test after several hours. I am sorry if I disappointed your enthusiasm. Thank You for the offer. –  user21436 Jan 29 '12 at 21:14