I was reading this question:
Let $N$ be a normal subgroup of index $m$. Prove that $a^{m}\in N$ for all $a\in G$.
Proof: Let $a \in G$. Since $[G:N] = m$, then $|G/N|=m$. From Lagrange's Theorem, it follows that $(aN)^m=a^mN=eN=N$. Hence $a^m\in N$.
For $|G/N|=m$ to follow from $[G:N]=m$, both $G$ and $N$ must be of finite order. But this assumption is not specified in the problem.
Is it possible to prove the same when $G$ and $N$ are of infinite order?
For example: Let us take $\mathbb{Z}$ under addition and $n\mathbb{Z}$ as a group and normal subgroup of infinite order. Then $[\mathbb{Z}:n\mathbb{Z}] = n$, so following the problem hypothesis, $x^n= nx\in n\mathbb{Z}$.
The problem statement seems to hold on that example. If the problem does not hold on the infinite example, how can I find a counterexample?
$G/N$ is the quotient group of $G$ mod $N$, where $N$ is a normal subgroup in $G$. The elements in $G/N$ are exactly the cosets of $N$ in $G$, so $|G/N|=[G:N]$. Given $[G:N]=m$, we know that $|G/N|=m$.
Note that $|G/N|$ is not $|G|/|N|$ unless both $G$ and $N$ are finite. The proof only cares that the quotient group $G/N$ is finite and doesn't really care if $G$ or $N$ are finite.
In your particular example, the quotient group $\mathbb{Z}/n\mathbb{Z} = \left\{n\mathbb{Z},n\mathbb{Z}+1,\ldots,n\mathbb{Z}+(n-1)\right\}$ is finite.