Cyclic Groups of Order p q/Proof 1
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
Extract the first part of this proof into a lemma that is then invoked in both this and Proof 2.
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Let $H$ be a Sylow $p$-subgroup of $G$.
Let $K$ be a Sylow $q$-subgroup of $G$.
By the Fourth Sylow Theorem, the number of Sylow $p$-subgroups of $G$ is of the form $1 + k p$ and divides $p q$.
We have that $1 + k p$ cannot divide $p$.
Then $1 + k p$ must divide $q$.
But as $q$ is prime, either:
- $1 + k p = 1$
or:
- $1 + k p = q$
But:
| \(\displaystyle 1 + k p\) | \(=\) | \(\displaystyle q\) | |||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle k p\) | \(=\) | \(\displaystyle q - 1\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle p\) | \(\divides\) | \(\displaystyle q - 1\) |
which contradicts our condition that $p$ does not divide $q - 1$.
Hence $1 + k p = 1$.
Thus there is only one Sylow $p$-subgroup of $G$.
Similarly, there is only one Sylow $q$-subgroup of $G$.
Thus, 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 $G$.
Why does it follow that $\order {x y} = \order x \order y = p q$?
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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:
| \(\displaystyle \order a\) | \(\divides\) | \(\displaystyle p\) | |||||||||||
| \(\, \displaystyle \land \, \) | \(\displaystyle \order a\) | \(\divides\) | \(\displaystyle q\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \order a\) | \(=\) | \(\displaystyle 1\) | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a\) | \(=\) | \(\displaystyle e\) |
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.
$\blacksquare$
