Showing that $f: \mathbb{Z} \to H$ is epimorphism implies $H$ is cyclic for negative $n$

Question

This is related to a question I asked here

At this point, I am specifically trying to show that a group $H$ is cyclic iff it is the homomorphic image of $\mathbb{Z}$. I have shown that $H$ cyclic $\implies$ is is the homomorphic image of $\mathbb{Z}$, and now I am attempting to show the other direction; however, I am at an impasse.

Thus far, I have the following:

Suppose that $f: \mathbb{Z} \to H$ is an epimorphism and let $a = f(1)$. Every $h \in H$ is of the form $h = f(n)$, $n \in \mathbb{Z}$.

If $n \geq 0$, then $h = f(1+ \cdots + 1) = f(1)\, \circ_{H}\, f(1)\, \circ_{H}\, \cdots \, \circ_{H} \, f(1) = (f(1))^{n} = a^{n}$ (where $\circ_{H}$ is the group operation in $H$).

For $n < 0$, I have been told the proof is very similar (so similar, in fact, that a lot of people don't even bother to prove it. I, however, have to.). However, I am not sure how to express 1) the sum of a negative number of $\mathbf{1}$'s in $\mathbf{\mathbb{Z}}$ and 2) a negative number of powers of $\mathbf{f(1)}$ in $\mathbf{H}.$ Without this, I am unable to finish the proof.

Could someone please show me how to do the thing I have outlined in bold face above (without asking me socratic questions, preferably. I am a very visual person and I need to see what this is supposed to look like.)?

Thank you.

Further Attempt: It can't be $h = f(-1-1-\cdots -1) =f(-1)\, \circ_{H}\, f(-1)\, \circ_{H}\, \cdots \, \circ_{H} \, f(-1) = (f(-1))^{n} = a^{-n} $, can it?

Answer 1

Summing a negative number of $1$'s doesn't make sense. You should instead sum a positive numbers of $(-1)$'s in $\Bbb Z$: $$-n=(-1)+(-1)+\dots +(-1).$$

Similarly, a negative power of $a$ in $H$ is in fact a positive power of the inverse of $a$, usually denoted $a^{-1}$: $$a^{-n}=(a^{-1})^n=a^{-1}\circ a^{-1}\circ \dots \circ a^{-1}.$$ Note that $f(-1)=a^{-1}$, since group homomorphisms preserve inverses. So you get that $$f(-n)=f((-1)+\dots +(-1))=f(-1)^n=(a^{-1})^n=a^{-n}$$ for all $n\geq 0$.

Answer 2

A group homomorphism sends zero to zero. So $f(0) = 0_H$. We have $f(1) + f(-1) = f(1+(-1)) = f(0) = 0_H$. This implies that $-f(1) = f(-1)$. Therefore $f(-n) = f\big((-1) + (-1) +\cdots + (-1)\big) = f(-1) + \cdots + f(-1) = n(-f(1))$.

Answer 3

Suppose $n=-m$, $m>0$. By definition of a group homomorphism, $$f(n)=f(-m)=\bigl(f(m)\bigr)^{-1}=\bigl(\bigl(f(1)\bigr)^{m}\bigr)^{-1}\stackrel{\text{def}}{=}\bigl(f(1)\bigr)^{-m}=\bigl(f(1)\bigr)^{n}$$