Group of Order p q is Cyclic/Proof 1
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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.
Proof
Lemma
There is:
- exactly one Sylow $p$-subgroup of $G$.
- exactly one Sylow $q$-subgroup of $G$.
$\Box$
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 $G$.
This article, or a section of it, needs explaining. In particular: Why does it follow that $\order {x y} = \order x \order y = p q$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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.
$\blacksquare$