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I've been looking into finding the presentation of subgroups $H$ of $G$ from a known presentation of the group $G$.
I know that usually, removing generators but keeping the set of relations the same can give presentations of subgroups. However, I've heard that there are instances where a subgroup might actually require more generators than the group itself. Does anyone know of any concrete examples of this that they can explain/know of where I can find an explanation?

David G. Stork
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anonymousgoose
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    This might interest you. – Shaun Mar 07 '18 at 23:08
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    This looks like an interesting discussion! Thanks for the link :) – anonymousgoose Mar 07 '18 at 23:08
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    It is not true that "removing generators but keeping the set of relations the same" usually gives a presentation of a subgroup. In fact usually you will still have relations that involve the removed generators, and so what remains will not be a presentation at all. – Derek Holt Mar 08 '18 at 08:16
  • Apologies that"s just what I'd heard; obviously the remaining relations would need to be altered to account for removed generators... – anonymousgoose Mar 09 '18 at 08:38

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This can happen even for finite groups: every finite group is a subgroup of a 2-generator group, namely a symmetric group. The simplest example of finite groups which requires more than two generators is $C_2 \oplus C_2 \oplus C_2$, where $C_2$ is cyclic of order 2. For infinite groups, it is a theorem Higman, Neumann and Neumann that every countable group is a subgroup of a group generated by two elements.

To find presentations of subgroups of a finitely presented group in practice, you may want to read about the Todd-Coxeter algorithm.

Robert Bell
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