Criterion for a group to be abelian Let $G$ be a group of order $pq$ , where $p,q$ are primes and $p<q$ . Show that if $p \nmid (q-1)$ (i.e  $q \not \equiv 1\mod p )$ then $G$ is abelian. 
I was able to use Sylow's theorems to prove that the group is cyclic and hence abelian (Sketch of what I did: the number of $p$-Sylow subgroups is $1$ or $q$  but since $q \not \equiv 1\mod p$ there is one $p$-Sylow subgroup and it is normal , similarly I showed that the number of $q$-Sylow subgroups is $1$ and also normal I then took their direct product, of the sylow subgroups that is, both are cyclic since they are of prime orders I then showed that the multiplication of generators generates the direct product which is isomorphic to $G$ it follows that G is abelian )
Besides the fact that the above proof looks very messy, I'm looking for a proof that does not use any Sylow theorem.  
 A: Yes, this can be done without Sylow as follows:
By Cauchy's theorem, the group has a subgroup of order $p$ and one of order $q$. If $p < q$ then the subgroup of order $q$ has index $p$ which is the smallest prime dividing the order of the group, so it is normal (see various proofs of this at Normal subgroup of prime index).
Now we know that the group is a semidirect product of $C_p$ with $C_q$ and since there is no non-trivial homomorphism from $C_p$ to $\rm{Aut}(C_q)$, since the latter has order $q-1$ which $p$ does not divide, this semidirect product is in fact direct, and we are done.
Just saw that you were not familiar with semidirect products, so here is a way without them:
We want to show that not only is the subgroup of order $q$ normal, it is in fact central. To see this, let $H$ be the $q$-Sylow subgroup (that normality implies uniqueness for Sylow subgroups is a lot easier to prove than the general theorems. For a generalization with an elementary proof, see Rotman introduction to theory of groups exercise).
We know from the N/C theorem that $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$ But we also know that since $H\leq C_G(H)$ we have that $|G/C_G(H)|$ divides $p$, and since $p$ does not divide the order of $\rm{Aut}(H)$ the only possibility is that $C_G(H) = G$ so $H$ is central.
Now let $K$ be any subgroup of order $p$. Since $|HK| = |G|$ we see that any element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$ and since $H$ is central, elements of this form commute, so the group is abelian.
Added: The N/C theorem is the following: Let $G$ be a group and $H$ a subgroup of $G$. Then $C_G(H)$ is a normal subgroup of $N_G(H)$ and $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$. The proof is by defining a homomorphism from $N_G(H)$ to $\rm{Aut}(H)$ by letting any $g\in N_G(H)$ go to the action on $H$ by conjugation and noting that this homomorphism has exactly $C_G(H)$ as its kernel.
A: By Cauchy's Theorem, $G$ contains subgroups of order $p$ and $q$. Let $x,y\in G$ such that $\langle x\rangle\cong\mathbb{Z}_p,\langle y\rangle\cong\mathbb{Z}_q$. Since $\mathbb{Z}_p\not\leq Aut(\mathbb{Z}_q)$, so $x$ centralises $y$, i.e., $x^{-1}yx=y$, and hence $x,y$ commute. Then $|\langle x,y\rangle|=pq$, and so $G=\langle x,y\rangle$ is abelian.
