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Let $G$ be the quotient group $\mathbb Q/\mathbb Z$. It is easy to show the following:

If $\mathcal C$ is a set of generators of $G$ (as $\mathbb Z$-module) and $x\in\mathcal C$, then $\mathcal C\setminus\{x\}$ also generates $G$.

In fact, let $x=\overline{m/n}\,.$ If $\ \overline{1/n^2}=\sum_{c\in\mathcal C}a_cc$, with $a_c\in\mathbb Z$, then $x=mn\,\overline{1/n^2}=mn(a_xx+\sum_{c\in\mathcal C\setminus\{x\}}a_cc)=\sum_{c\in\mathcal C\setminus\{x\}}a_cmnc$ (because $nx=0$).

The same method of proof can be tried in order to obtain a similar result for $\mathbb Q$ instead of $G$; however I get stuck at some point:

Let $\mathcal C$ be a set of generators of $\mathbb Q$ as $\mathbb Z$-module, and let $x\in\mathcal C$, say $x=m/n$. We have $1/n^2=\sum_{c\in\mathcal C}a_cc$, with $a_c\in\mathbb Z$, so $x=mn(1/n^2)=mn(a_xx+\sum_{c\in\mathcal C\setminus\{x\}}a_cc)=a_xm^2+\sum_{c\in\mathcal C\setminus\{x\}}a_cmnc$.

If $x\ne1$ and $1\in\mathcal C$, we are done; otherwise I cannot imagine how to circunvent this exceptional case. Can we remove any single generator always, or there are exceptions?

Matemáticos Chibchas
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2 Answers2

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We can remove any single generator, yes.

In any group $G$, an element $x\in G$ is called a non-generator if it can safely be removed from any generating set. The set of all non-generators is actually a subgroup, called the Frattini subgroup, and it is the intersection of all maximal subgroups of $G$.

Here, it is understood that if $G$ has no maximal subgroups then the Frattini subgroup is $G$ itself. You should be able to prove that $(\mathbb{Q},+)$ indeed has no maximal subgroups, and therefore every element including $1$ is a non-generator.

Alon Amit
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    It is not understood, but rather it is what the OP is trying to prove. – Pedro Nov 04 '16 at 02:38
  • Not sure I follow. I just highlighted what happens to the definition of the Frattini subgroup in the absence of maximal subgroups. I didn't attempt to prove the result he's trying to prove - he's quite close anyway. I just answered his question, and provided context. – Alon Amit Nov 04 '16 at 02:39
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I managed to solve my problem, although the solution is somewhat "devious". Any more direct solution will be welcomed. Until then, I will reluctantly accept my own answer ;-).

Let $\mathcal C$ be a set of generators of $\mathbb Q$ as $\mathbb Z$-module. Let $x\in\mathcal C$, and let $\mathcal C'=\mathcal C\setminus\{x\}$. We have $x/2=ax+s$, with $a\in\mathbb Z$ and $s\in\langle\mathcal C'\rangle$, so $(1-2a)x\in\langle\mathcal C'\rangle$. Now $x/(1-2a)=bx+t$, with $b\in\mathbb Z$ and $t\in\langle\mathcal C'\rangle$, hence $x=b(1-2a)x+(1-2a)t\in\langle\mathcal C'\rangle$, as desired.

Matemáticos Chibchas
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