## There is a unique group homomorphism from ZZ to any group where the image of 1 is fixed

Show that if $H$ is a group and $h \in H$, there exists a unique homomorphism $\varphi : \mathbb{Z} \rightarrow H$ such that $\varphi(1) = h$.

Existence: Define $\varphi(n) = h^n$. We need not worry about well-definedness for $\varphi$. For all $m,n \in \mathbb{Z}$, we have $\varphi(m+n) = h^{m+n}$ $= h^mh^n = \varphi(m)\varphi(n)$, so that $\varphi$ is a homomorphism.

Uniqueness: Suppose we have another homomorphism $\psi$ such that $\psi(1) = h$. Then $\psi(n) = \psi(n \cdot 1)$ $= \psi(1)^n = \varphi(1)^n$ $= \varphi(n)$, so that $\psi = \varphi$ and so $\varphi$ is unique. (Keep in mind that we write $Z$ additively and $H$ multiplicatively.)

• Bobby Brown  On September 14, 2010 at 1:54 pm

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?

• nbloomf  On September 14, 2010 at 2:11 pm

It is because we’re thinking of $\mathbb{Z}$ as an additive group. There, the identity is 0 and “powers” are written as “multiples”. $H$, 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!)

• Bobby Brown  On September 14, 2010 at 4:24 pm

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?