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If a group and subgroup have a finite index, and we know that the subgroup is finitely generated, can we conclude that the group is also finitely generated?

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closed as off-topic by user26857, 6005, user99914, user91500, tired Nov 2 '15 at 12:43

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  • $\begingroup$ Sounds likely., $\endgroup$ – Gregory Grant Nov 1 '15 at 1:20
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Let $H=\langle h_1,\ldots, h_n\rangle$ and let $[G:H]=m$. Suppose $g_1,\ldots,g_m$ is a complete set of coset representatives for $G/H$. Since, $$G=\bigcup_{k=1}^m g_kH$$ we can see that $G=\langle\{h_jg_k:1\le j\le n,1\le k\le m\}\rangle$.

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  • $\begingroup$ This is exactly what I needed to understand. The only typo (unless I am missing something) is that m and n are transposed at the end. $\endgroup$ – Jill_Johnson Nov 1 '15 at 1:32
  • $\begingroup$ Yes, you are right about that. I'll fix it. :) $\endgroup$ – Tim Raczkowski Nov 1 '15 at 1:34

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