Why not embed F/R into F?

1. Aug 28, 2008

tgt

Why doesn't such an embedding exist? When R is a normal subgroup of F.

2. Aug 28, 2008

morphism

Why would such an embedding exist? Conceptually it doesn't make sense.

By way of example, can you embed Z/2Z into Z?

3. Aug 28, 2008

tgt

Sure, I see the exampe but can you explain the conceptual part? How is it conceptual to you?

4. Aug 28, 2008

morphism

Well, if you think about it for a bit, what does it mean to be able to embed F/R into F? Intuitively, modding out by a subgroup means you collapse that subgroup to zero. So for F/R to be isomorphic to some subgroup of F, it would be necessary for R to be 'complemented' inside of F by another subgroup (think of complementation here as you would in the setting of vector spaces). I personally don't see any reason why one would believe every normal subgroup of F to have this property. (But I admit to being prejudiced: functional analysis has made me very suspicious of the process of complementation!)

To make this a bit more precise, we can use semidirect products: If F/R embeds into F, then F is essentially $R \rtimes F/R$. It's easy to see that a necessary and sufficient condition for this to happen is that the canonical exact sequence

$$1 \longrightarrow R \longrightarrow F \longrightarrow F/R \longrightarrow 1$$

splits.

By the way, is there any specific reason you're using F and R to denote groups?!

5. Aug 28, 2008

tgt

Some good points raised. Although what does split mean?

Those notations naturally fit into the notation of a presentation.

Last edited: Aug 28, 2008