Show that if is a group and , there exists a unique homomorphism such that .

Existence: Define . We need not worry about well-definedness for . For all , we have , so that is a homomorphism.

Uniqueness: Suppose we have another homomorphism such that . Then , so that and so is unique. (Keep in mind that we write additively and multiplicatively.)

## Comments

In this one, I am having trouble understanding why:

psi(n*1)=psi(1)^n

can anyone give a bit more explanation to this step?

It is because we’re thinking of as an additive group. There, the identity is 0 and “powers” are written as “multiples”. , on the other hand, is written multiplicatively. I can see why this might be confusing- I’ll add a bit of explanation.

(Thanks for reading, by the way!)

No I understand that part. Ah I see now though…you meant psi(1+1+…+1), with 1 appearing n times, then you can split that into psi(1)*…*psi(1), also appearing n times, and so on. Thanks and you’re welcome.

I study math at Colorado (I was accepted to Arkansas, but opted for here…) and this has been a great algebra resource for me. Do you know of something similar to this in analysis, perhaps with Royden’s book?

Good luck with your studies

I’m glad to hear you find this useful. Unfortunately I don’t know of any similar resources for analysis.

Good luck to you as well.