Sylow p-Subgroups of Group of Order 2p/Proof 1
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Theorem
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.
Proof
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the Fourth Sylow Theorem:
- $n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
- $n_p \divides 2 p$
that is:
- $n_p \in \set {1, 2, p, 2 p}$
But $p$ and $2 p$ are congruent to $0$ modulo $p$
So:
- $n_p \notin \set {p, 2 p}$
Also we have that $p > 2$.
Hence:
- $2 \not \equiv 1 \pmod p$
and so it must be that $n_p = 1$.
It follows from Sylow $p$-Subgroup is Unique iff Normal that this Sylow $p$-subgroup is normal.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(1)$ Groups of order $2p$