Let $G$ be the group defined by the presentation $\langle a\ | \ a^{n}=e\rangle$ and $n>0$. I want to prove that $G$ is finite.
Set $H=\{x\ |\ (\exists k)(k\in\mathbb{Z}\ \&\ x=a^k)\}$. Then $H=G$. We already know that there exists a surjective homomorphism $f:\mathbb{Z}\rightarrow G$ defined by $f(k)=a^k$ for $k\in\mathbb{Z}$. Thus there exists a subgroup of $\mathbb{Z}$ of the form $m\mathbb{Z}$ for $m\geq0$ such that $\tilde{f}:\mathbb{Z}/m\mathbb{Z}\rightarrow G$ is an isomorphism. This shows that $|G|=[\mathbb{Z}:m\mathbb{Z}]$.
Now, how can I conclude that $n=m?$ What goes wrong if I assume $m=0$? In that case $G$ would be isomorphic to $\mathbb{Z}$, but does that give me a contradiction?
I guess what I am having difficulty with is this: how does $a^n=e$ force $G$ to be finite?
We can show directly that $G = \langle a \, | \, a^n = e \rangle = \{e, a, \dots, a^{n - 1}\}$. Indeed, for any integer $k$, we may write $k = nq + r$ for some integer $q$ and $0 \leq r < n$. Then $a^k = a^{nq + r} = a^{nq}a^r = e^qa^r = a^r \in G$. Hence $G$ is finite.
If we equip $G$ with the obvious operation, then $G$ becomes (isomorphically) $\mathbb{Z} / n\mathbb Z$.