Why doesn't this presentation embed?

[tgt]

How to show <G,t|t^{-1}kt=k, k in K> does not embed into <G,t|->?

Where K is a subgroup of an arbitrary group G.

 

[morphism]

I admit I don't know much about this stuff, but what exactly does <G,t|-> mean? Is it the free group generated by the set $G \cup \{t\}$? And how are you defining the embedding of presentations?

 

[tgt]

I admit I don't know much about this stuff, but what exactly does <G,t|-> mean? Is it the free group generated by the set $G \cup \{t\}$? And how are you defining the embedding of presentations?

If G=<X|R> where X is the generating set for G and R are the relations in G then <G,t|-> is <X,t|R>.

Embedding means there exists an injective homomorphism.

 

[morphism]

I still don't understand. What is "-", the empty relation? If so then isn't what I asked true: <G,t|-> is the free group generated by $G \cup \{t\}$?

 

[tgt]

I still don't understand. What is "-", the empty relation? If so then isn't what I asked true: <G,t|-> is the free group generated by $G \cup \{t\}$?

I never said you were wrong. I like to only use the term free group for groups generated from a basis only. What I said in the previous post has no mistakes either.

 

[tgt]

Correction, you were actually wrong. The free group generated by G U {t} will be greater. i.e take G={a,b} with the relations a=1, bb=1. Those relations will hold in <G,t|-> but not in the free group. In fact <G,t|-> is the free product of G with {t}.