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Is the following statement true: Any surjective homomorphism $f:G\to H$ of groups with equal rank maps every minimal generating system $x$ of $G$ to a minimal generating system $y$ of $H$? "Minimal generating set" means "generating set of minimal cardinality". Is it only valid for finite rank? How about infinite rank?

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    $\begingroup$ (1) Could you replace "a" with either "some" or "every" to clarify the question? (2) (after Igor's comment): minimal means, not properly containing any generating subset. There are often minimal generating subsets that don't achieve the rank (e.g., $S_n$ generated by $n-1$ transpositions while it's also 2-generated). If you mean "of minimal cardinal", please correct. $\endgroup$
    – YCor
    Commented Dec 19, 2017 at 3:38

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I'm assuming that "minimal generating set" means "a generating set that does not properly contain a generating set" (that is, every proper subset does not generate). In that reading, the statement is false in both finite and infinite ranks.

For a counterexample in finite rank, let $G$ be infinite cyclic generated by $x$ (rank $1$), and $H$ be cyclic of order $2$ generated by $y$ (also rank $1$). Let $f\colon G\to H$ be the map sending $x$ to $y$. The minimal generating set $\{x^2,x^3\}$ (which generates $G$, but no proper subset does) is mapped to $\{e,y\}$, which is not minimal. Or you could take $H$ to be cyclic of order $4$, if you want to avoid the trivial element, as then you get $y^2$ and $y^3$; which is not minimal.

For infinite rank, take $G=H$ be a direct sum of countably infinitely many copies of the infinite cyclic group; take $\{x_i\}_{i=1}^{\infty}$ as a basis for $G$, and $\{y_j\}_{j=1}^{\infty}$ as a basis for $H$. Take $f\colon G\to H$ be the map that sends $x_{2i-1}$ to $y_i$ and $x_{2i}$ to $y_i^{-1}$.


If by "minimal generating set" you mean "generating set of minimal cardinality", then given that the rank of a group is the least cardinality of a generating set, then the answer is positive: under a surjective homomorphism, a generating set is mapped to a generating set. So if $X$ is a generating set for $G$ with $\mathrm{rank}(G)$ elements, then $f(X)$ is a generating set for $H$ with at most $\mathrm{rank}(G)=\mathrm{rank}(H)$ elements, and since any generating set has at least $\mathrm{rank}(H)$ elements, it follows that $f(X)$ has exactly $\mathrm{rank}(H)$ elements (in cardinality) and so is a generating set of minimal cardinality.

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  • $\begingroup$ It is not clear that this is what the OP means by "minimal". He could mean "of mininal cardinality". $\endgroup$
    – Igor Rivin
    Commented Dec 19, 2017 at 3:09
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    $\begingroup$ @IgorRivin: If that's the case, then the answer is trivially yes, since generating sets map to generating sets under surjective maps. $\endgroup$ Commented Dec 19, 2017 at 15:36
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    $\begingroup$ @ArturoMagidin you also interpreted "a" as "every". I'm still waiting for clarification from the OP. $\endgroup$
    – YCor
    Commented Dec 19, 2017 at 16:53
  • $\begingroup$ I am not arguing that point. $\endgroup$
    – Igor Rivin
    Commented Dec 19, 2017 at 16:54
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    $\begingroup$ @YCor: The second part would show that "there exists a minimal generating set (whether of minimal cardinality or minimal in not properly containing a generating set) that is mapped to a minimal generating set" holds. $\endgroup$ Commented Dec 19, 2017 at 17:31

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