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

Question

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 $q$?

EDIT: Thank you for your answers, I will look into it.
I haven't learned about Sylows theorems or groups yet. Is there any other way to prove this without using Sylow?

Answer 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.

Answer 2

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.

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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.

contradiction.

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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 $q \mid s \mid p-1$.

Answer 3

Assume the existence of a subgroup of order $p$ by Cauchy's Theorem. Let us call it $H$ and $a$ be its generator. This must be normal from this answer. So for any element $b$ in $G,$ we get that $b^{-1}ab = a^d$ for some integer $d$ not divisible by $p.$ Then

$$b^{-(p-1)}ab^{p-1} = b^{-(p-2)}a^d b^{p-2} = b^{-(p-3)} ( b^{-1}ab )( b^{-1}ab ) \dots ( b^{-1}ab ) b^{p-3} = b^{-(p-3)} a^{d^2} b^{p-3} = a^{d^{p-1}} = a,$$ because $p$ divides $d^{p-1} - 1,$ for $d.$ This is a consequence of Lagrange's Theorem.

So the subgroup generated by $b^{p-1}$ commutes with $a.$ If $b^{p-1}$ is not the identity then $G$ has to be abelian.

Answer 4

You can address the contrapositive claim, namely:

If $q\nmid p-1$, then $G$ is abelian.

By cardinality reasons, there is one subgroup of order $p$, only, hence normal in $G$. Moreover, there is a subgroup of order $q$ (corollary of Cauchy's theorem). Therefore, $G\cong C_q\ltimes C_p$. But $\left|\operatorname{Aut}(C_p)\right|=$ $p-1$ and $C_q$ is simple; if $q\nmid p-1$, then $C_q$ cannot embed into $\operatorname{Aut}(C_p)$, and we're left with the trivial action, only. So, $G\cong C_q\ltimes C_p=$ $C_q\times C_p\cong$ $C_{pq}$, which is abelian (cyclic, indeed).