13
$\begingroup$

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.

Thanks for your help!

$\endgroup$
24
$\begingroup$

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 $n \in \mathbb N$.

$\endgroup$
  • 13
    $\begingroup$ Just a little comment for spohreis's attention. Note that Kannappan has proved that no divisible group has a subgroup of finite index. $\endgroup$ – user23211 Jan 28 '12 at 14:49
2
$\begingroup$

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 $\mathfrak m$.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – user21436 Jan 29 '12 at 15:10
  • $\begingroup$ 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. $\endgroup$ – Pierre-Yves Gaillard Jan 29 '12 at 16:22
  • $\begingroup$ 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. $\endgroup$ – user21436 Jan 29 '12 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.