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Mr Davis 97

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Mr Davis 97

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That it is a group.Suppose that we know that ##G=\langle S \mid R\rangle##, that is, ##G## has a presentation. If ##N\trianglelefteq G##, what can be said about ##G/N##?

A relation is a word with letters from the set of generators which multiplies to ##1##. Now ##\pi\, : \,G \twoheadrightarrow G/N## is a group homomorphism, so ##\pi(R)=1_{G/N}## where the left hand side is a word with letters from the images of the generators. Say ##R=a^nb^m## then ##N=(a^nb^m)N=(aN)^n(bN)^m##. You could say that ##\bar{R}=\pi(R)## is a relation in ##G/N##.I know that for example, if ##G=\langle x,y \rangle##, then ##G/N = \langle xN, yN \rangle##. But is there anything that can be said about the relations in ##G##as they might correspond to relations for ##G/N##, or is there no correspondence in the relations?

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