# $(\mathbb{Q},+)$ has no maximal subgroups

I have a problem that I don't have any idea.

Show that group $(\mathbb{Q},+)$ has no maximal subgroups.

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What's a maximal subgroup? – Rudy the Reindeer Nov 11 '12 at 17:06
No maximal proper subgroup, I think. What does a subgroup look like? – Mark Bennet Nov 11 '12 at 17:08
definition maximal subgroup is proper subgroup – Muniain Nov 12 '12 at 1:25

Suppose $H$ is any nonzero proper subgroup of $\mathbb Q$ and let $x \in \mathbb Q \setminus H$ and $y \in H, y \neq 0$

Write $\dfrac {y}{x} = \dfrac {a}{b}$ with integers $a,b$. Then $a \neq 0$ and $\dfrac {x}{a} \notin H + \langle x \rangle$ : Suppose $\dfrac {x}{a} = h + nx$ for some $n \in \mathbb Z$ and $h \in H$. Then $x = ah+anx = ah+nby \in H$, which contradicts the hypothesis on $x$. Thus $H$ is not maximal.

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Are you assume that $H$ is maximal subgroup of $\mathbb{Q}$? – Muniain Nov 12 '12 at 9:13
@Firmino : no I pick any proper subgroup and show that it is not maximal. – mercio Nov 12 '12 at 9:27
and use $H+\langle x \rangle = \mathbb{Q}$ to implies $x/a \notin \mathbb{Q}$, contradicts?. Thank you so much. – Muniain Nov 12 '12 at 16:18
@Firmino : Since $x/a \notin H+\langle x \rangle = H'$, $H'$ is a proper subgroup of $\mathbb Q$, and since $x \in H'$ and $x \notin H$, $H$ is strictly smaller than $H'$, thus $H$ isn't a maximal proper subgroup. – mercio Nov 12 '12 at 19:06
I think $H+\langle x \rangle =\mathbb{Q}$. If reverse then we have $H+\langle x \rangle$ is proper subgroup of $\mathbb{Q}$, implies $H$ is proper subgroup of $\mathbb{Q} - \langle x \rangle$. contradict with maximal property of $H$. – Muniain Nov 13 '12 at 11:05

Fixed it

Assume by contradiction that $H$ is a maximal subgroup of $\mathbb Q$.

As for $r \neq 0$ the function $f(x)=rx$ is a group automorphism of $\mathbb Q$, by replacing $H$ by $f(H)$ we can assume without loss of generality that $1 \in H$.

Now, if $\frac{1}n \in H$ for each $n > 1$ it is easy to prove that $H =\mathbb Q$. Pick $n$ to be the smallest $n$ such that $\frac{1}{n} \notin H$.

Then $H + < \frac{1}{n} > =\mathbb Q$.

Now, for each positive integer $l$ if $l=qn+r$ we have $\frac{l}{n}=q+\frac{r}{m}$ and $q \in H$.

It follows from the above that each rational number can be written in the form $$r=h+\frac{k}{n} \, \mbox{ with } h \in H, 0 \leq k < n$$

Therefore, $$\frac{1}{n^2}= h+\frac{k}{n} \, \mbox{ with } h \in H, 0 \leq k < n$$

Multiplying both sides by $n$ we get $$\frac{1}{n}= nh+k \in H$$ as $nh \in H$ and $k \in h$.

Are you suggesting that for any subgroup $H$ and any element $x \notin H$, then $x/2 \notin H+<x>$ ? because that's not true. – mercio Nov 11 '12 at 17:19
+1. No contradiction is required btw. Your argument shows that any subgroup that is not $\mathbb{Q}$ is contained in a larger subgroup that is also not $\mathbb{Q}$ and can therefore not be maximal. – WimC Nov 11 '12 at 17:20
Well, pick $H = \langle \frac 1 2 \rangle$ and $x = \frac 1 3$. I'm pretty sure $\frac 1 6 = \frac 1 2 - \frac 1 3$ – mercio Nov 11 '12 at 17:40
@dh87 Because $H \subset H+<x> \subset \mathbb Q$ and $x \in H+<x>, x \not in H$. This implies that $H \neq H+<x>$. – N. S. Mar 30 '15 at 23:44
The group $\mathbb Q$ is a divisible group. There is a well known facts that says $G$ is divisible if and only if $G$ has no maximal subgroups if and only if every nonzero quotient of $G$ is infinite.