Here is the problem:
Suppose $(G, .)$ is a finite Abelian group and has the property that $\forall k \in N, |\{ g ∈ G : g^k = e \}| ≤ k$. Show that G is cyclic.
And this is my lecturer's proof:
Let $n = |G|$. It is enough to show that $G$ has an element of order $n$. If $g \in G, g \ne e$ then it's order $d$ divides $n$. Moreover $H = \langle g \rangle$ contains all elements of order $d$. As $H$ is a cyclic group of order $d$, it follows that the number of elements of order $d$ in $H$ (and therefore in $G$) is $\phi(d)$. Thus, the number of elements of $G$ of order $d$ is $\phi(d)$. Since $\sum_{d|n} \phi(d) = n$ if $d|n$, then number of elements of $G$ of order $d$ is $\phi(d)$. In particular, there are $\phi(n)$ elements of $G$ of order $n$. As $\phi(n) \ne0$, $G$ is therefore cyclic.
What I'm having trouble understanding is where in the proof the fact that $G$ is Abelian and the property is used. In other words where does this argument fall apart for any general group.