Group of Order p q is Cyclic

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Theorem

Let $p, q$ be primes such that $p < q$ and $p$ does not divide $q - 1$.

Let $G$ be a group of order $p q$.

Then $G$ is cyclic.

Lemma

There is:

exactly one Sylow $p$-subgroup of $G$.

exactly one Sylow $q$-subgroup of $G$.

Proof 1

By Sylow $p$-Subgroup is Unique iff Normal, $H$ and $K$ are normal subgroups of $G$.

Let $H = \gen x$ and $K = \gen y$.

To show $G$ is cyclic, it is sufficient to show that $x$ and $y$ commute, because then:

$\order {x y} = \order x \order y = p q$

where $\order x$ denotes the order of $x$ in $G$.

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Since $H$ and $K$ are normal:

$x y x^{-1} y^{-1} = \paren {x y x^{-1} } y^{-1} \in K y^{-1} = K$

and

$x y x^{-1} y^{-1} = x \paren {y x ^{-1} y^{-1} } \in x H = H$

Now suppose $a \in H \cap K$.

Then:

\(\ds \order a\)

\(\divides\)

\(\ds p\)

\(\, \ds \land \, \)

\(\ds \order a\)

\(\divides\)

\(\ds q\)

\(\ds \leadsto \ \ \)

\(\ds \order a\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds e\)

where $e$ is the identity of $G$.

Thus:

$x y x^{-1}y^{-1} \in K \cap H = e$

Hence $x y = y x$ and the result follows.

$\blacksquare$

Proof 2

Let the Sylow $p$-subgroup of $G$ be denoted $P$.

Let the Sylow $q$-subgroup of $G$ be denoted $Q$.

We have that:

$P \cap Q = \set e$

where $e$ is the identity element of $G$.

Hence in $P \cup Q$ there are $q + p - 1$ elements.

As $p q \ge 2 q > q + p - 1$, there exists a non- identity element in $G$ that is not in $H$ or $K$.

Its order must be $p q$.

Hence, by definition, $G$ is cyclic.

$\blacksquare$

Also see

• Group Direct Product of Cyclic Groups: a similar result which can often be confused with this one.

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \alpha$