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I'm trying to prove that if $H$ is a normal subgroup of a group $G$ such that $H$ and $G/H$ are finitely generated, then G is finitely generated also. I'm trying to find a finite set $X$ such that $G$ is generated by $X$, but I have no ideal how to find this set using the finite generator sets of $H$ and $G/H$.

I need help

Thanks in advance

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  • $\begingroup$ Can you do this if instead of groups they are vector spaces? $\endgroup$ – Mariano Suárez-Álvarez Apr 28 '13 at 20:43
  • $\begingroup$ @MarianoSuárez-Alvarez I think yes $\endgroup$ – user42912 Apr 28 '13 at 20:58
  • $\begingroup$ Well: exactly the same argument works for groups. $\endgroup$ – Mariano Suárez-Álvarez Apr 28 '13 at 20:59
  • $\begingroup$ But what is $G/H$ in vector spaces algebra? $\endgroup$ – user42912 Apr 28 '13 at 21:01
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    $\begingroup$ If you don't now what that quotient is for vector spaces, how can you do this for vector spaces? $\endgroup$ – Mariano Suárez-Álvarez Apr 28 '13 at 21:05
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Hints: we're given

$$H=\langle\,h_1,\ldots,h_k\,\rangle\;,\;\;G/H:=\langle\,g_1H,\ldots,g_nH\,\rangle$$

Remember now that for all $\,x\in G\,$ there exist unique $\,1\le i_x\le n\,$ and unique $\,h_x\in H\,$ s.t. $\,x=g_{i_x}h_x\,$ and, of course, then $\,x\in g_{i_x}H\,$ , so...

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  • $\begingroup$ How does the normality of H come into play? $\endgroup$ – user41442 Apr 29 '13 at 0:10
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    $\begingroup$ Otherwise you have no quotient group at all,@user41442... $\endgroup$ – DonAntonio Apr 29 '13 at 1:44

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