Need a big list of proves of proper subgroups of Q2 to be cyclic.

Question

Here is the proposition:

Let Q$_2$:={$\frac{a}{b}\in\mathbb Q:a,b\in\mathbb Z$ and $b=2^n$, where n is non-negative integer}. Then Q$_2$ is a group under addition. If $H$ is a proper subgroup of Q$_2$ such that $\mathbb Z\subseteq H$, then $H$ is cyclic.

Notice that Q$_2$ is not cyclic, since there always has a $l$ such that group generated by $\frac{a}{2^k}$, $a\in \mathbb Z$ and $k\in \mathbb Z^+\cup\{0\}$ does not include element $\frac{1}{2^l}$ where $l\gt k$.

In the other words, Q$_2$ is the smallest subgroup of itself that contains $\mathbb Z$ and is not cyclic.

I have my proof through elementary group theory, but I think it could be proved in others ways, perhaps through Zorn's lemma. Also, my instructor says my proof is not neat, there is an easy way to prove it.

Therefore, I want to have a big list of the proves. You are welcome to use every kinds of math tools. Category theory, number theory...

I will post my proof just after below the description.

Answer 1

Here is a short proof.

Consider the set $$S=\{\frac{1}{2^k}\ |\ k\ge0,\ \frac{1}{2^k}\in H\}$$ Then notice $\mathbb Z\in H$, so $1\in H\Rightarrow\ S\neq\emptyset$

Now, we claim that $S$ must be finite if $H$ is proper subgroup of $ \mathbf Q_2$. Suppose not, then $\forall \ k\ge 0,\ \frac{1}{2^k}\in H$. Since $H$ is a addictional group, we have $\forall \ k\ge0, \forall a\in \mathbb Z, \frac{a}{2^k}\in H$, i.e. $H=\mathbf Q_2$. Contradicts $H$ is a proper subgroup of $\mathbf Q_2$. Hence, $S$ is finite.

Then, $\exists\ b\ge 0$ s.t. $\frac{1}{2^b}$ is the smallest element in $H$. Claim $H=\langle\frac{1}{2^b}\rangle$. Suppose not, then $\exists \ x\in H, s.t. x\notin\ \langle\frac{1}{2^b}\rangle$. Then wen could write $x=\frac{2k+1}{2^{b+n}}$ for some $k,\ n\in \mathbb Z$ and $n\ge1$. Then $2^{n-1}x\in H\Rightarrow\frac{2k+1}{2^{b+1}}\in H$. We know that $\frac{-k}{2^b}\in H\Rightarrow\frac{-k}{2^b}+\frac{2k+1}{2^{b+1}}=\frac{1}{2^{b+1}}\in H$ which contradicts the minimality of $\frac{1}{2^b}$. Therefore, $H$ is cyclic.

Q.E.D

Answer 2

See, if for every $n\in \mathbb{N}$ if there exists an element $k/2^n \in H$ then $H=Q_2$, as if $k/2^n$ is there then $k/2^m$ is also there for all natural number $m<k$, and since all the integers are also there so $1/2,1/2^2\dots 1/2^k$ are also there, hence the generator of the whole group $Q_2$ will also be there.

So there exists an $N\in \mathbb{N}$ such that $k/2^N$ is not there for any natural number $k$, Now define $M=max\{n\in \mathbb{N}: 1/2^n \in H\}$, clearly $1/2^M$ this generates $H$.

Answer 3

First step, we claim that any subgroup of $(1/2^n)\mathbb Z:=\{\frac{p}{2^n}\in\mathbb Q:p\in\mathbb Z\}$ contains $\mathbb Z$ Is in form $(1/2^k)\mathbb Z$, where $k\in \mathbb Z,0\le k\le n$.

Proof of the claim:

Let $H$ be a subgroup of $(1/2^n)\mathbb Z$ (Or in notation $H\le (1/2^n)\mathbb Z$). Notice that the subgroups of $(1/2^n)\mathbb Z$ must in form $(1/2^n)m\mathbb Z$ for some $m\in \mathbb Z$.

In fact $H$ is a subgroup of $(1/2^n)\mathbb Z$ iff $H=$$(1/2^n)m\mathbb Z$. Since ($\Leftarrow$) if $H=(1/2^n)m\mathbb Z=\{\frac{mp}{2^n}:p\in\mathbb Z\}$, then $\frac{mp}{2^n}\in(1/2^n)\mathbb Z$, i.e. $H\le (1/2^n)\mathbb Z$. Conversely ($\Rightarrow$) if $H\le(1/2^n)\mathbb Z$, then pick the smallest positive element of $H$, say $a$. Then $a\in (1/2^n)\mathbb Z$, i.e. $\exists\ q\in \mathbb Z$ s.t. $a=\frac{q}{2^n}$, or we can use $\frac{q}{2^n}$ to represent the smallest positive element. Suppose there is a p s.t. $\frac{p}{2^n}\neq \frac{kq}{2^n},\forall\ k\in \mathbb Z$, then $\frac{p}{2^n}=\frac{kq+r}{2^n}$ where $0\le r\lt q$, then $\frac{r}{2^n}=\frac{p-kq}{2^n}\in H$, which indicates that $\frac{r}{2^n}\lt \frac{q}{2^n}$. This contradicts the fact that $\frac{q}{2^n}$ is the smallest element in $H$. Hence the subgroups of $(1/2^n)\mathbb Z$ are in form $(1/2^n)m\mathbb Z$ for some $m$.

Now, back to trip, since $H$ Is a subgroup of $(1/2^n)\mathbb Z$, then $H=(1/2^n)m\mathbb Z$ for some $m\in \mathbb Z$. Since $\mathbb Z\le H=(1/2^n)m\mathbb Z$ i.e. $\exists\ p\in\mathbb Z$ s.t. $\frac{mp}{2^n}=1$, or $mp=2^n$, then $m=2^k$, $p=2^{n-k}$ for some $0\le k \le n$. Then we have $H=(1/2^n)2^k\mathbb Z=(1/2^{n-k})\mathbb Z$, or $H=(1/2^k)\mathbb Z$, where $k\in\mathbb Z,\ 0\le k\le n$.

Second step, we prove the proposition. We first use the claim above. $H\le$Q$_2$ and $\mathbb Z\le H$, if $\exists q$ s.t. $H\le(1/2^n)\mathbb Z$, then $H$ Is cyclic.

Proof:

Let $H\le$ Q$_2$, s.t. $\mathbb Z\le H$. If $\exists \ q$, s.t. $H\le (1/2^q)\mathbb Z$, then by the above claim, we know that $H$ is cyclic. Else, assume that $H$ is not cyclic. By the contrpositve of the claim, if $H$ Is not cyclic, then $\forall q\ge0,q\in \mathbb Z$, $H$ is not the subgroup of $(1/2^q)\mathbb Z$. THat is to say, for a fixed q, $H$ is not the subgroup of $(1/2^q)\mathbb Z$, where $q$ can be any non-negative integer. Then for the fixed $q$, $H$ is not the subgroup of $(1/2^q)\mathbb Z$, i.e. $\exists h\in H$, s.t. $h\notin (1/2^q)\mathbb Z$. However, $h\in$Q$_2$, since $H\le$Q$_2$. Then $h=\frac{a}{2^γ}$, for some integer $a$ and $γ\ge 0$. Since $\frac{a}{2^\gamma}\notin (1/2^q)\mathbb Z$,$\Rightarrow\ \gamma \gt q$, then $H\ge \langle\frac{a}{2^\gamma}\rangle$(the group generated by $\frac{a}{2^\gamma}$). Notice $\mathbb Z\le H$, then $H\ge \mathbb Z+(a/2^\gamma)\mathbb Z=\frac{(2^\gamma+a)\mathbb Z}{2^\gamma}=\frac{gcd(2^\gamma,a)\mathbb Z}{2^\gamma}$. Notice, $1\le gcd(2^\gamma,a)\le \gamma -q-1$(since $\frac{a}{2^\gamma}\notin\frac{\mathbb Z}{2^q}$, then $a=$ odd number$\times 2^l$, where $l\in \{0,1,2,...,\gamma-q-1\}$). Therefore, $H\ge \frac{2^{\gamma-q-1}\mathbb Z}{2^\gamma}=\frac{\mathbb Z}{2^{q+1}}\ge \frac{\mathbb Z}{2^q}$. Since q is arbitrary, i.e. $H\supseteq\{\frac{a}{2^q}:a,q\in\mathbb Z,q\ge 0\}$,or $H\supseteq$Q$_2$. Since, Q$_2$ is a group, and $H\le$Q$_2$, we have $H=$Q$_2$, this contradicts that $H$ Is a proper subgroup of Q$_2$. Hence, $H$ is cyclic.

Q.E.D

Answer 4

Assume that Q$_2$ is cyclic. Then Q$_2$ must be isomorphic to $\mathbb{Z}$ since it is infinite.

However, this is not possible because Q$_2$ has the property that for every element $a$ there exists an element $b$ such that $a=b+b$; whereas $\mathbb{Z}$ contains many elements without that property.

Answer 5

To start, let just recall that a group is cyclic if and only if it is generated by one element, and the group generated by an element $x$ (written in additive notation) has the form $\langle x\rangle=\{m\cdot x:m\in\mathbb{Z}\}$, where

• $m\cdot x:=\underbrace{x+\cdots +x}_{m-times}$ if $m>0$.
• $m\cdot x:=-(\,\underbrace{x+\cdots +x}_{m-times}\,)$ if $m<0$.
• $0\cdot x=0_H$.

In particular, we have the following easy claim, which I highlight as it will be used several times.

Claim 1: Let $H$ be a group. If $x\in H$, then $a\cdot x\in H$ for all $a\in\mathbb{Z}$.

This is very straightforward but I add a proof now: since $a\in\mathbb{Z}$, $|a|\in \mathbb{N}$ and since $H$ is a subgroup of $\textbf{Q}_2$ it is closed under addition and (additive) inverse, so we have $$a\cdot x:=sign(a)(\,\underbrace{x+\cdots+x}_{|a|-times}\,)$$

[Here $sign(a)$ denotes the sign of $a$.]

Now, let $H$ be a subgroup of $\textbf{Q}_2$ containing $\mathbb{Z}$.

Claim 2: If $\frac{1}{2^n}\in H$ for all $n\in\mathbb{N}$, then $H=\textbf{Q}_2$.

This follows easily from Claim 1: By assumption $H$ is a subgroup of $\textbf{Q}_2$, hence $H\subseteq \textbf{Q}_2$. For the other contenence, let $x=\dfrac{a}{2^n}$ be an element in $\textbf{Q}_2$. Then, $a\in\mathbb{Z}$ and by the hypothesis of Claim 2 we have that $\frac{1}{2^n}\in H$. Thus, by Claim 1, $$x=\dfrac{a}{2^n}=a\cdot \frac{1}{2^n}\in H$$

This shows that $\textbf{Q}_2\subseteq H$, completing the proof of Claim 2.

Proposition: Suppose that $H$ is a proper subgroup of $\textbf{Q}_2$. Then $H$ is cyclic.

By Claim 2, since $H\subsetneq \textbf{Q}_2$, there is a minimal $m\in\mathbb{N}$ such that $\frac{1}{2^{m+1}}\not\in H$. We will show that $H$ is precisely the subgroup of $\textbf{Q}_2$ generated by $\frac{1}{2^m}$.

First, notice that $\frac{1}{2^m}\in H$ by the minimality of $m$, and since $H$ is a group, the subgroup generated by $\frac{1}{2^m}$ is contained in $H$. Conversely, suppose for a contradiction that there is $n\in\mathbb{N}$ minimal such that $x=\dfrac{a}{2^n}\not\in \langle \frac{1}{2^m}\rangle$ for some integer $a$.

If $\frac{1}{2^n}\in H$, we would have by Claim 1 that $x\in H$. Also, we must have $gcd(a,2)=1$, since otherwise we can reduce the fraction to obtain a smaller $n$ with the same property.

Hence, there are integers $x,y$ such that $ax+2y=1$. By the minimality of $n$ we have that $\frac{1}{2^{n-1}}\in H$, and by Claim 1 we have

$$\frac{1}{2^n}=\frac{ax+2y}{2^n}=x\cdot\frac{a}{2^n}+y\frac{1}{2^{n-1}}\in H,$$ contradiction.

Answer 6

Define $G_n=\frac{1}{2^n}\mathbb{Z}=\{\dfrac{a}{2^n}:a\in\mathbb{Z}\}$. Notice that, by definition, $\textbf{Q}_2=\bigcup_{n\geq 0} G_n$, and each of the subgroups $G_n$ is cyclic.

If $H$ is a proper subgroup of $\textbf{Q}_2$ containing $\mathbb{Z}=G_0$, there is a maximal $m\in\mathbb{N}$ such that $G_m\subseteq H$. If $H\setminus G_m\neq \emptyset$, we can choose a minimal $n>m$ such that $x=\frac{a}{2^n}\in H\setminus G_m$ for some $a\in\mathbb{Z}$. By the minimality of $n$, we have that $a$ is odd, say $a=2b+1$ for some $b\in\mathbb{Z}$.

Given that $\frac{a}{2^n}\in H$, then $(2^{n-(m+1)})\cdot \frac{a}{2^n}=\frac{a}{2^{m+1}}\in H$, and since $G_m\subseteq H$, also $\frac{b}{2^m}\in H$. Thus, we have $\frac{1}{2^{m+1}}=\frac{(2b+1)-(2b)}{2^{m+1}}=\frac{a}{2^{m+1}}-\frac{b}{2^m}\in H$, which implies that $G_{m+1}=\langle \frac{1}{2^{m+1}}\rangle\subseteq H$, contradicting the maximality of $H$.

Therefore, $H=G_m$, and this shows that $H$ is cyclic.