In a comment to Arturo Magidin's answer to this question, Jack Schmidt says that the additive group of the rationals has no minimal generating set.
Why does $(\mathbb{Q},+)$ have no minimal generating set?
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4 Answers
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3 Let $S$ be a minimal generating set of $\mathbb{Q}$, and take $a\in S$. Let $H\leq \mathbb{Q}$ be the subgroup generated by the elements of $S$ which aren't $a$. Because $S$ is minimal, $H$ is a proper subgroup of $\mathbb{Q}$. Define $G:=\mathbb{Q}/H$. Then $G$ is non-trivial and cyclic (it is generated by the class of $a$), so $G\cong\mathbb{Z}/n\mathbb{Z}$ or $G\cong \mathbb{Z}$. This is a contradiction, because every element of $\mathbb{Q}$ is divisible by $n$, so the same should be true of a quotient of $\mathbb{Q}$.
- $\begingroup$ Why $H\cap\langle a\rangle=0$? $\endgroup$Camilo Arosemena Serrato– Camilo Arosemena Serrato2013-09-08 23:26:16 +00:00Commented Sep 8, 2013 at 23:26
- $\begingroup$ I've allowed for the possibility that $H\cap\langle a\rangle = \langle a^n\rangle$. In this case $G\cong \mathbb{Z}/n\mathbb{Z}$. $\endgroup$Julian Rosen– Julian Rosen2013-09-08 23:28:06 +00:00Commented Sep 8, 2013 at 23:28
- $\begingroup$ ok, now I see... $\endgroup$Camilo Arosemena Serrato– Camilo Arosemena Serrato2013-09-08 23:29:06 +00:00Commented Sep 8, 2013 at 23:29
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Let $S\subseteq\mathbb Q$ be such that $\langle S\rangle=\mathbb Q$. Fix $a\in S$, and put $T=S\setminus\{a\}$, let us see that also $\langle T\rangle=\mathbb Q$. We have $$\frac{a}{2}=a\cdot k_0+\sum_{i=1}^na_i\cdot k_i,$$ for some $k_i\in\mathbb Z$ and $a_i\in T$. Then $$a=a\cdot (2k_0)+\sum_{i=1}^na_i\cdot (2k_i),$$ that is, $$a\cdot m=\sum_{i=1}^na_i\cdot (2k_i),$$ where $m=1-2k_0$ is nonzero; as $k_0$ is an integer.
Now $\frac{a}{m}$ can be expressed as a combination of elements of $S$, say $\frac{a}{m}=a\cdot r_0+\sum_{i=1}^lb_i\cdot r_i,$ with $b_i\in T$,$r_i\in\mathbb Z$, thus $$a=a\cdot mr_0+\sum_{i=1}^lb_i\cdot mr_i=\sum_{i=1}^na_i\cdot r_0(2k_i) +\sum_{i=1}^lb_i\cdot mr_i.$$
answered Sep 8, 2013 at 23:04
Camilo Arosemena Serrato
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factor out the integers, and consider $Q_*$ the additive rationals modulo unity.
each prime $p$ may be associated with a 'pure' subgroup $Q_{*p}$containing all (and only) the fractions whose denominator is a power of $p$. such a subgroup is the limit by set union of an ascending chain of subgroups generated by a monotone sequence of (negative) powers of $p$, for example:
$$ <\frac{1}{p}>\; \subset \;<\frac{1}{p^2}> \; \subset \dots $$ any cofinal sequence will do, but any non-cofinal (i.e. finite, here) sequence must exclude denominators with a power of $p$ greater than the maximal power in the finite chain.
thus it is trivial that for a pure subgroup no generator is indispensible.
but $Q_*$ is the direct sum of its pure subgroups. so by an easy argument, no generator can be indispensible.
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I was struggling with this problem yesterday and now I found a solution which is similar to Camilo's. I will use the same notation. $$a=a\cdot k_0+\sum_{i=1}^na_i\cdot k_i,$$ for some $k_i\in\Bbb{Z}$ and $a_i\in T$. Then $$a(1-k_0)=\sum_{i=1}^na_i\cdot k_i.$$ Letting $b_i=a_i/(1-k_0)$ -- clearly if $a_i$ generate $\Bbb{Q}$ then $b_i$ also generate it. Also we can assume that $a\not=b_i$ for any $i$ -- we get $$\sum_{i=1}^nb_i\cdot k_i=\dfrac{1}{1-k_0}\sum_{i=1}^na_i\cdot k_i=a.$$
