Existence of a minimal generator for a group? Let $G$ be a group and $A\subseteq G$ and $G=\left<A\right>$. Is there a minimal $B\subseteq A$ with $G=\left<B\right>$?
 A: If you are not quite satisfied with $\Bbb Q$ under addition, you can also take the Prufer $p$-group for some prime $p$. The Prufer $p$-group is usually realized as a subset the unit circle of the complex plane under complex multiplication, as seen here.
That a generating set for this group doesn't contain minimal generating set lends itself to be more intuitive than $\Bbb Q$; at least, I think so.
As an explicit example let's take the generating set $A=\left\{\exp\left(\frac{2\pi i}{p^k}\right)\right\}_{k=1}^\infty$. It should be clear that no finite subset of this set will generate $G$, for the generated set will then simply be $\left\{\exp\left(\frac{2\pi im}{p^n}\right)\right\}_{m\in\Bbb Z}$ where $n$ is the greatest exponent of $p$ occurring in our finite subset. In fact, every finitely-generated subset of the Prufer $p$-group is cyclic. So if a minimal generating set existed, it would have to be countably-infinite.
But we can remove any element of the form $\exp\left(2\pi i/p^k\right)$ from this generating set, because the subgroup generated by $\exp\left(2\pi i/p^n\right)$ with $n>k$ will account for that element we removed. So we can remove any finite subset of this generating set. We can say a little more. We can even remove any collection of the form $\{\exp\left(2\pi i/p^k\right)\}$ as long as for each $k$ we remove there is a $n>k$ such that $\exp\left(2\pi i/p^n\right)$ is still in our remaining subset. For example, we can remove all of the elements of the form $\exp\left(2\pi i/p^k\right)$ with $k$ even (or $k$ odd).
To make it precise, assume we chiseled out the perfect subset $S$ of the natural numbers (starting at 1) such that $\{\exp\left(2\pi i/p^k\right)\}_{k\in S}$ was a minimal generating subset of $G$ contained in $A$. Either $S$ is countable or finite. But by our previous discussion, $S$ cannot be finite otherwise it generates a finite set. And if $S$ is countable, let $n$ be the first element of $S$. Then $\{\exp\left(2\pi i/p^k\right)\}_{k\in S-\{n\}}$ is also a generating set, contradicting the definition of $S$.
A: Non-artinian modules with infinitely many generators will give you a counter example then( not all of them but thats where you check first). like @MJD said $\langle \Bbb Q, +\rangle$ gives us counter example.
