6
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

contradiction.


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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.