I have a problem that I don't have any idea.
Show that group $(\mathbb{Q},+)$ has no maximal subgroups.
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Martin Sleziak
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Muniain
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What's a maximal subgroup? – Rudy the Reindeer Nov 11 '12 at 17:06
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No maximal proper subgroup, I think. What does a subgroup look like? – Mark Bennet Nov 11 '12 at 17:08
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9definition maximal subgroup is proper subgroup – Muniain Nov 12 '12 at 01:25
4 Answers
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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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1@Firmino : no I pick any proper subgroup and show that it is not maximal. – mercio Nov 12 '12 at 09:27
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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
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4@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
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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
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what's $\mathbb Q - \langle x \rangle$ ? $H + \langle x \rangle$ isn't $\mathbb Q$ because $x/a$ is not in it and I haven't supposed $H$ maximal. But yes if you really need to show the claim by contradiction, you can add "suppose $H$ is maximal" at the beginning and "This contradicts our assumption that $H$ was maximal" at the end. – mercio Nov 13 '12 at 15:56
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I can see the proof is correct but I can't find why it happens. Could you explain the idea behind these computations? Please. – Lord Shadow Nov 13 '20 at 16:50
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I feel like your choice of x/a has something to do with powers of prime in denominator. But I can't be sure. – Lord Shadow Nov 13 '20 at 17:06
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@LordShadow It's easy to just check that $H+\langle x \rangle$ is a subgroup of $\mathbb{Q}$ that contains $H$. – Bcpicao Jan 12 '21 at 18:23
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There is another simple proof, maybe an explanation of this answer:
Assume that $N$ is a maximal subgroup of $\mathbb{Q}$, then the quotient group $\mathbb{Q}/N$ must be simple abelian group, i.e. cyclic group with prime order. Let's say $\mathbb{Q}/N \cong \mathbb{Z}_p$ for some prime $p$. Since $N \neq \mathbb{Q}$, there exists $a \notin N$. Since $\mathbb{Q}/N \cong \mathbb{Z}_p$, we have $N = p(\frac{a}{p} + N) = a + N$ which implies $a \in N$, a contradiction.
Zelox
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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 $k \in h$.
This is a contradiction.
Martin Sleziak
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N. S.
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2Are you suggesting that for any subgroup $H$ and any element $x \notin H$, then $x/2 \notin H+
$ ? because that's not true. – mercio Nov 11 '12 at 17:19 -
2+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
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2Well, 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
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I'm sorry for bringing up the old question. Just a quick question: Why do we have $H$ is maximal implies $H+
=\mathbb{Q}$? – dh16 Mar 30 '15 at 23:05 -
1@dh87 Because $H \subset H+
\subset \mathbb Q$ and $x \in H+ – N. S. Mar 30 '15 at 23:44, x \not in H$. This implies that $H \neq H+ $.
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The group $\mathbb Q$ is a divisible group. There is a well known fact 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.
Martin Sleziak
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Mikasa
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