Group of order 15 is abelian How do I prove that a group of order 15 is abelian?
Is there any general strategy to prove that a group of particular order (composite order) is abelian?

(Without: Sylow, Cauchy, semi-direct products, cyclic $G/Z(G)$ argument, $\gcd(n,\phi(n))=1$ argument. Just Lagrange and the class equation.)
Firstly, if $|G|=pq$, with $p,q$ distinct primes, say wlog $p>q$, then $G$ can't have $|Z(G)|=p,q$, because otherwise there's no way to accomodate the centralizers of the noncentral elements between the center and the whole group (recall that, for all such $x$, it must strictly hold $Z(G)<C_G(x)<G$).
Next, if in addition $q\nmid p-1$, then a very simple counting argument suffices to rule out the case $|Z(G)|=1$. In fact, if $Z(G)$ is trivial, then the class equation reads:
$$pq=1+kp+lq \tag 1$$
where $k$ and $l$ are the number of conjugacy classes of size $p$ and $q$, respectively.  Now,  there are exactly $lq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$). Since each subgroup of order $p$ comprises $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $lq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because by assumption $q\nmid p-1$).
Therefore, $(1)$ yields:
$$pq=1+kp+m'q(p-1) \tag 2$$
for some positive integer $m'$; but then $q\mid 1+kp$, namely $1+kp=nq$ for some positive integer $n$, which plugged into $(2)$ yields:
$$p=n+m'(p-1) \tag 3$$
In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$, but this is not essential here). So, $1+kp=q$: contradiction, because $p>q$.
So we are left with $|Z(G)|=pq$, namely $G$ abelian (and hence, incidentally, cyclic).
