A group of order $pq$ I want to prove that if a group of order $pq$ (where $p,q$ are primes) with $p>q$ is not abelian, then $p \equiv1$ mod $q$. I don't know if this is correct but I think I have a proof using characters: 
$G$ is abelian if and only if $G'=[G,G]$ is trivial. Otherwise, it should be $|G'|=p$ or $|G'|=q$ or $|G'|=pq$. Let's see what happens.
If $|G'|=p$, then $|G/G'|=q$ and therefore there are $q$ linear characters of $G$. Hence $\sum_{\chi\in Irr(G)}\chi(1)^2=|G|$ implies that $q+\sum_{\chi\in Irr(G),\chi(1)\neq 1}\chi(1)^2=pq$. We know that $\chi(1)|pq$ and so if $\chi(1)\neq 1$ it should be $\chi(1)=p$ or $\chi(1)=q$ or $\chi(1)=pq$. Since $q<p$, it is easily seen that for all $\chi$  (that are not of degree 1) we necessarily have $\chi(1)=q$. Therefore if $N$ is the number of characters with $\chi(1)=q$, we get $q+Nq^2=pq$ and therefore $p=Nq+1$, i.e $p \equiv 1$ mod $q$. With very similar reasoning we see that it cannot happen that $|G'|=q$ of $|G'|=pq$ and so if $G$ is not abelian, then $p \equiv1$ mod $q$.
Is the above correct? And if so, can you find a proof without using characters (maybe that's easy, but I really can't see it..)? 
 A: Your answer is correct. Here is a more "standard" approach, using the Sylow-theorems (which should be the first thing you think of given your assumptions):
By Sylow's third theorem we have a unique normal Sylow $p$-group $P$ (use $1 < q < p)$.
Consider the quotient $G/P$. This quotient is cyclic as it has prime order and thus abelian. Therefore, the commutator $G'$ is contained in $P$. I.e. $G' \subseteq P$. Because $G$ is non-abelian, we have more than $1$ Sylow $q$-group (otherwise you can show that the unique Sylow q subgroup $Q$ contains the commutator $G'$ and then it will follow that $G' =1$, which means that $G$ is abelian). Thus $|Syl_q(G)| = p$ (this order must divide the order of $G$). However, $p =|Syl_q(G)| \equiv 1\bmod q$ and your result follows.
Alternatively, if you want to avoid the commutator subgroup, you can argue by contradiction that $|Syl_q(G)| =1$, but then there is a normal subgroup $Q$ of order $q$ and then it follows that $G \cong P \times Q$, so that $G$ is abelian. Contradiction.
A: I think that a solution based solely on the Class Equation can be given, as follows.
For your group, the Class Equation reads (see e.g. here):
$$pq=1+k_pp+k_qq \tag 1$$
where $k_i$ are the number of conjugacy classes of size $i$.  Now,  you have exactly $k_pp$ elements of order $q$ (they are the ones in the conjugacy classes of size $p$)$^\dagger$. Since each subgroup of order $q$ contributes $q-1$ elements of order $q$, and two subgroups of order $q$ intersect trivially, then $k_pp=l(q-1)$ for some positive integer $l$ such that $p\mid l$ (because $p\nmid q-1$). Therefore $(1)$ yields:
$$pq=1+l'p(q-1)+k_qq \tag 2$$
for some positive integer $l'$; but then $p\mid 1+k_qq$, namely $1+k_qq=np$ for some positive integer $n$, and finally $(2)$ yields:
$$q=n+l'(q-1) \tag 3$$
which holds for arbitrary $q$ if and only if $l'=1$ and $n=1$, whence in particular $1+k_qq=p$ or, equivalently, $p\equiv 1 \pmod q$.
Incidentally, this provides also an answer to this other question, where the roles of $p$ and $q$ are swapped, being there $p<q$. In fact, the number of conjugacy classes is given by:
\begin{alignat}{1}
|G/\sim_{conj}| &= 1+k_p+k_q \\
&= 1+(q-1)+\frac{p-1}{q} \\
&= q+\frac{p-1}{q} \\
\end{alignat}

$^\dagger$ An element lies in a conjugacy class of size $p$ (respectively, $q$) if and only if its order is $q$ (respectively, $p$). This claim follows from the Orbit-Stabilizer Theorem and the fact that, for $g\in G\setminus\{e\}$, it is $\langle g\rangle=C_G(g)$.
