# Show that every proper subgroup of this group is finite.

Let $G$ be the group of rational numbers in $[0,1)$ whose denominator is a power of $2$: \begin{align*} G &= \{r/2^k : \text{$r \in \mathbb Z$, $0 \le r < 2^k$, $k = 0, 1, \ldots$} \} \\ &=\{0, \frac12, \frac14, \frac34, \frac18, \frac38, \frac58,\frac78, \frac1{16}, \ldots \} \end{align*}

Addition in $G$ is modulo $1$. So $3/4 + 5/8 = 3/8$. Show that every proper subgroup of $G$ is finite.

I was planning to define $A_k = \{r/2^k : r = 0,1, \ldots, 2^k - 1\}$.

Then it is not hard to show that $A_k$ is a subgroup of $G$,

then, $A_k \subseteq A_{k+1}$, where $A_k$ is a cyclic group of order $2^k$, and $G = \bigcup_{k=0}^\infty A_k$.

Since every $A_i$ is finite, I am done.

Not sure if I'm overthinking too much, feel like this question is harder than a few lines. Am I missing something here?

• You did not prove that every subgroup is finite, you just exhibited a family of finite subgroups whose union is $G$. There could be other infinite subgroups. – Denis Sep 23 '14 at 9:08
• $[ 0 , 1 )$ with addition modulo $1$ is isomorphic to the multiplicative group of complex numbers of absolute value $1$, that is the unit circle. $G$ is isomorphic to the subgroup of roots of unity of order a power of $2$. The result follows from the fact that every finite subgroup of the multiplicative group of a field is finite. – lhf Sep 23 '14 at 10:59
Hint: Try to show that if a subgroup of $G$ is infinite, then for infinitely many $k$ it contains a generator of $A_k$. Then show that it is in fact true for all $k$.
Hint: Let $H$ be a proper subgroup of $G$. Let $\frac{2j+1}{2^k}\in G-H$. Now argue that no reduced fraction in $G$ with denominator greater than or equal to $2^k$ is in $H$. Thus all denominators in $H$ (when fractions are expressed in reduced form) are less than $2^k$. How many such fractions are there?