# Natural group homomorphism

1. Jan 16, 2008

### samkolb

1. The problem statement, all variables and given/known data
Let G and G' be groups. Let H and H' be normal subgroups of G and G' respectively. Let
f:G->G' be a group homomorphism.

Show that if f(H) is a subset of H', then there exists a natural homomorphism f*:(G/H)->(G'/H'). (Hint: This fact is used constatntly in algebraic topology.)

2. Relevant equations
If X is a normal subgroup of a group Y, then the factor group Y/X can be formed where the group operation is coset multiplication.

3. The attempt at a solution
Using the assumptions that H is normal in G, I showed that f(H) is a normal subgroup of f(G).

I formed the factor group f(G)/f(H) and defined a function f*:(G/H)->(f(G)/f(H)) by
f*(aH) = (f(a))(f(H)) which turns out to be a homomorphism.

This seems like a "natural" thing to do, so I think I'm on the right track. The problem is that my function f* is not mapping to the right place: I don't think that f(G)/f(H) is a subset of G'/H'.

I also noticed that in proving the above, I did not make use of the hypotheses that H' is normal in G' or that f(H) is a subset of H'. Can these assumptions be used to show that
f(H) = H'?

Any hints would be greatly appreciated.
Sam

2. Jan 16, 2008

### Mathdope

Admittedly, I haven't put anything down on paper here, but does a problem arise if you define your function as f*(aH) = f(a)H' ?

3. Jan 16, 2008

### morphism

Well, you are on the right track. Try drawing a diagram. We have 3 maps: f:G->G', and the two natural projection maps G->G/H and G'->G'/H'. Try to make the diagram commute.