# Finite non-Abelian group $|G| = pq$, $p>q$ primes, prove: $q\ |\ p-1$

I have a finite noncommutative group $$G$$ with $$pq$$ elements, where $$p, q$$ are prime numbers.
So $$|G| = pq$$ and $$p > q$$.
I need to prove that $$p-1$$ is divisible by $$q$$. (so that $$q\ |\ p-1$$)
I think I am supposed to use centralizers. (Centralizers for element $$a \in G$$ is a set $$R(a) = \{g^{-1}ag\ |\ g \in G\}$$.)

I have proved that there exist one and one only subgroup with $$p$$ elements and that there are $$p-1$$ elements with order $$p$$ in the group $$G$$. I am not sure if this is useful.
How could I prove that $$p-1$$ is divisible by $$q$$?

I haven't learned about Sylows theorems or groups yet. Is there any other way to prove this without using Sylow?

Suppose $$q$$ does not divide $$p-1$$. Denote the number of Sylow $$q$$-subgroups of $$G$$ by $$n_q$$. According to the Sylow's theorem, $$n_q \mid p$$ and $$n_q \equiv 1$$ mod $$q$$. Since $$q \not \mid p-1$$ we have $$n_q \neq p$$. Thus $$n_q=1$$.

Combining this with your result, we can prove $$G$$ is abelian. Write the unique Sylow $$p$$-subgroup and $$q$$-subgroup by $$P$$ and $$Q$$, respectively. They are normal subgroups of $$G$$. By Lagrange's theorem, $$|P \cap Q|=1$$. Hence $$|PQ|=|P||Q|=pq=|G|$$. It follows that $$G=P \times Q \cong \mathbb{Z}_p \times \mathbb{Z}_q \cong \mathbb{Z}_{pq}$$, which is cyclic.

EDIT

I haven't learned about Sylows theorems or groups yet. Is there any other way to prove this without using Sylow?

Here is one without Sylow's theorem.

I have proved that there exist one and one only subgroup with $$p$$ elements and that there are $$p−1$$ elements with order $$p$$ in the group $$G$$.

Denote the unique subgroup of order $$p$$ by $$P$$. For given $$g \in G$$, the conjugation map $$x \mapsto g^{-1}xg$$ is an group automorphism. Thus $$|g^{-1}Pg|=p$$ and so we obtain $$g^{-1}Pg=P$$. This shows $$P$$ is normal.

Since $$G$$ is nonabelian, there is no element $$x \in G$$ of order $$pq$$. By Lagrange's theorem, any element of $$G$$ has order $$1, q, p, pq$$. Hence there are exactly $$pq-p$$ elements of order $$q$$. As a result, there exists at least one subgroup $$Q$$ of $$G$$ of order $$q$$.

Now $$PQ$$ is a subgroup of $$G$$ since $$P \unlhd G$$ and $$Q \leq G$$. By comparing cardinalities, we obtain $$G=PQ$$. (By Lagrange $$|P \cap Q|=1$$.) Let $$x$$ and $$y$$ be generators of $$P$$ and $$Q$$, respectively. From the normality of $$P$$ we have $$y^{-1}xy = x^k$$ for some $$1 \leq k < p$$.

We need an observation:

If $$z\in Q$$ satisties $$xz=zx$$, then $$z$$ must equal to $$e$$, the identity element of $$G$$.

Suppose $$z \neq e$$ satisfies $$xz=zx$$. Then $$Q$$ is generated by $$z$$ so $$G=PQ$$ is generated by $$\{x, z\}$$. Since $$z$$ commutes with generators, $$z$$ must lie in the center of $$G$$. But $$Z(G)$$ is trivial because of the following proposition:

If $$G/Z(G)$$ cyclic, then $$G$$ is abelian.

Now come back to our situation. We know $$y^{-1}xy = x^k$$ holds for some $$1 \leq k < p$$. Consider $$\mathbb{Z}_p=\{ \overline{0}, \overline{1}, ..., \overline{p-1} \}$$ and its unit group $$\mathbb{Z}^{\times}_p=\{\overline{1}, ..., \overline{p-1} \}$$. It is known that $$\mathbb{Z}^{\times}_p$$ is cyclic.. Denote the order of $$\overline{k}$$ in $$\mathbb{Z}^{\times}_p$$ by $$s$$. By the Lagrange's theorem $$s$$ divides $$p-1$$.
Computations show that $$y^{-2}xy^2 = y^{-1}x^ky = (y^{-1}xy)^k=x^{k^2}$$ $$y^{-3}xy^3 = y^{-1}x^{k^2}y=(y^{-1}xy)^{k^2}=x^{k^3}$$ Inductively, we have
$$y^{-s}xy^s = x^{k^s}=x$$ The last equality holds since $$k^s \equiv 1$$ mod $$p$$. Therefore $$y^s=e$$ and $$q \mid s$$. Thus $$q \mid s \mid p-1$$.
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.