Show that there exists a unique homomorphism $f: \mathbb{Z} \to G$ such that $f(1)=a$

Question

Here, $a$ is any element of the group $G$.

I'm not really sure, intuitively, what this proposition is really saying. I understand the idea that there exists a homomorphism from $\mathbb{Z}$ to any group $G$, but why does $1\in \mathbb{Z}$ always have to be mapped to an element of $G$? The only thing I can think of is that $\langle 1 \rangle = \mathbb{Z}$, and so $f(n) = n \cdot f(1)$ for any integer $n$. Since $a$ is the image of $1$, and $\langle 1 \rangle = \{1^n : n\in \mathbb{Z}\}$ (where $1^n = n\cdot 1$ because $\mathbb{Z}$ is a group under $+$), then this might have some kind of relationship with $f(n) = n \cdot f(1)$. I mean, if we just replace $f(1)$ with $a$, then we get $f(n) = n\cdot a$, which is remarkably similar to $\langle 1 \rangle$. I think I'm pretty close, but I'm struggling to make a couple connections.

Answer 1

Note that $\Bbb Z$ is a cyclic group generated by $1$.

If $f:\Bbb Z\to G$ is any homomorphism then $f(n)=f(1+1+\ldots(\text{n times})+1)=nf(1)$.

So once you fix $f(1)$ the whole homomorphism is fixed and hence is unique for a particular choice of $f(1)\in G$

Answer 2

Yes you are right. Since 1 generates $\mathbb{Z}$ you can define the homomorfism $f$ as $f(1)=a$ and $f(n)= f(1+1+...+1)= a+a+...+a=na$ and that's all you need!. we have now determined the value of every integer and it is indeed a homomorphism since $f(n+m)= (n+m)a=na+ma=f(n)+f(m)$. Actually the image of $f$ will be the ciclic subgroup of $G$ generated by $a$. I hope I've answered your question usefully.