Subgroup of Order p in Group of Order 2p is Normal/Corollary

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Theorem

Let $p$ be an odd prime.

Let $G$ be a group of order $2 p$ whose identity element is $e$.

Let $a \in G$ be of order $p$.

Let $K = \gen a$ be the subgroup of $G$ generated by $a$.


Let $G$ be non-abelian.

Every element of $G \setminus K$ is of order $2$, and:

$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$


Proof

By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$.


$1$ is not possible because Identity is Only Group Element of Order 1.

Then we have that $G$ is non-abelian.

Hence from Cyclic Group is Abelian, $G$ is not cyclic.

Thus $\order b \ne 2 p$.


It remains to investigate $2$ and $p$.


Let $b \in G \setminus K$.

By Subgroup of Index 2 contains all Squares of Group Elements:

$b^2 \in K$

Aiming for a contradiction, suppose $b$ has order $p$.

Then by Intersection of Subgroups of Prime Order:

$K \cap \gen b = e$

which contradicts $b^2 \in K$.

Thus by Proof by Contradiction $\order b \ne p$


Hence it must follow that $\order b = 2$.

$\Box$


We have that:

$b a \in G \setminus K$

Thus:

$\paren {b a}^2 = e$
\(\ds b a\) \(\in\) \(\ds G \setminus K\)
\(\ds \leadsto \ \ \) \(\ds \paren {b a}^2\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds b a b\) \(=\) \(\ds \paren {b a}^2 a^{-1}\)
\(\ds \) \(=\) \(\ds a^{-1}\)
\(\ds \leadsto \ \ \) \(\ds b a b^{-1}\) \(=\) \(\ds a^{-1}\) as $b = b^{-1}$

$\blacksquare$


Sources

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