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Let $G$ be a finite group possessing a generating set of order $n \in \mathbb{N}$. Let $H \leq G$ and $x_1, \dots, x_n \in G$ for which $\langle H, x_1, \dots, x_n \rangle = G$. Must there be $h_1, \dots, h_n \in H$ such that $\langle x_1h_1, \dots, x_nh_n \rangle = G$?

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up vote 6 down vote accepted

Here's an example. Let $S$ be a simple nonabelian group, $s$ an element of order 2 in $S$ and $c\in S$ such that $S$ is generated by $s$ and $c$ (there are plenty of examples).

Let $G=S^3$; it can be generated by 2 elements. Define $H=\{(1,1,1),(1,s,s),(s,1,s),(s,s,1)\}$. It is easy to check that the subgroup generated by $H$ and $(c,c,c)$ is all of $S^3$: indeed its projection to each pair of factors is onto, hence it contains one of the factors, and since it's invariant par permutation of coordinates it contains all factors.

Now let $x_1=(c,c,c)$ and $x_2=(1,1,1)$, so that $S^3=\langle H,x_1,x_2\rangle$. I claim there are no $h_1,h_2$ in $H$ such that $S^3=\langle x_1h_1,x_2h_2\rangle=\langle x_1h_1,h_2\rangle$. Fix $h_1,h_2$, define $L=\langle x_1h_1,h_2\rangle$. Up to permuting coordinates, we can suppose that $h_2$ is either $(1,1,1)$ or $(s,s,1)$. Then the projection of $L$ to the 3rd coordinate is cyclic; hence $L\neq S^3$.

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Do you think there can be a reason to believe that this holds true (in the topological sense) if $G$ is a free finitely generated profinite group? – Pablo Jul 19 '14 at 21:16
My example gives an example with the free profinite group on 2 generators $PF_2$: just take a continuous homomorphism from $PF_2$ onto $S^3$, define the new $H$ as the pull-back of $H\subset S^3$, and $x_1,x_2$ any lifts of the old $x_1,x_2$. – YCor Jul 19 '14 at 21:24
I have some feeling that if $G$ is solvable, then this could be true. What do you think? – Pablo Jan 3 '15 at 16:03
Then I think you should try to prove (or disprove) it. – YCor Jan 3 '15 at 17:32

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