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If $G$ can be generated by n elements, $A$ is a subgroup of $G$ and index $(G:A)$ is finite,

I was required to prove: $A$ can be generated by $2n(G:A)$ elements.

There is an answer that A generally cannot be generated by only n elements when G is non-abelian, but the upper bound here is much higher.(If a group $G$ is generated by $n$ elements, can every subgroup of $G$ by generated by $\leq n$ elements?)

Thank you!

Micheal Johnson
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  • In fact $(n-1)|G:A|+1$ elements are enough to generate $A$. This follows from the result for free groups - see here for example. – Derek Holt Jan 13 '21 at 08:02
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    Why the close votes? There are many questions around that are for more worthy of close votes than this one. – Derek Holt Jan 13 '21 at 08:07
  • The group $G$ is not necessarily free. It is finitely generated but can have any relation between the generators, thus $[G:A]$ may be smaller than $[F(S):A]$ when S generates G. – Micheal Johnson Jan 13 '21 at 08:22
  • But the point is that the result for $G$ follows immediately from the result for the free group $F$ on $n$ generators, because $G$ is a quotient group of $F$. – Derek Holt Jan 13 '21 at 09:10

1 Answers1

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Welcome to MSE!

It seems this answer does what you want. In particular, the factor of $2n$ comes from ensuring the generating set $Y$ of $G$ symmetric, and the factor of $(G:A)$ comes from taking a transversal $S$ of $G/A$.

Following that answer, if we write $[g]$ to mean the chosen representative of $Ag$ in $S$, induction on word length shows $\{ sy [sy]^{-1} ~|~ s \in S, y \in Y \}$ (which has size $2n[G:A]$) generates $A$.

I hope this helps ^_^

Chris Grossack
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    Followed your link and found an elegant proof (albeit took me some time to understand it). Thank you! – Micheal Johnson Jan 13 '21 at 07:38
  • But the question does not ask for a symmetric generating set, so $n|G:A|$ generators suffice (which can be improved to $(n-1)|G:A|+1$). – Derek Holt Jan 13 '21 at 08:03
  • I don't think it's true, if Y isn't symmetric (after careful analyzing) we may well find ${ sy^{-1} [sy^{-1}]^{-1} ~|~ s \in S, y \in Y }$ among the generating elements of A. I hope others to clarify though. – Micheal Johnson Jan 13 '21 at 08:13
  • But those generators are redundant, because $sy^{-1}[sy^{-1}]^{-1} = ([sy^{-1}]ys^{-1})^{-1} = ([sy^{-1}]y[sy^{-1}y]^{-1})^{-1}$ (hope I've got that right!) – Derek Holt Jan 13 '21 at 09:12