Cauchy's Group Theorem/Proof 1
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Theorem
Let $G$ be a finite group whose identity is $e$.
Let $p$ be a prime number which divides order of $G$.
Then $G$ has a subgroup of order $p$.
Proof
Let $\order G$ be a prime number.
Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$.
Now suppose $e \ne g \in G$.
Let $\order g = n$.
Let $p \divides n$.
Then by Subgroup of Finite Cyclic Group is Determined by Order the cyclic group $\gen g$ has an element $g$ of order $p$.
Suppose $p \nmid n$.
From Subgroup of Abelian Group is Normal, $\gen g$ is normal in $G$.
Consider the quotient group $G' = \dfrac G {\gen g}$.
As $p \nmid n$, $\gen e \subsetneq \gen g \subsetneq G$.
Thus $\order {G'} < \order G$.
But we have that $p \divides \order G$.
It follows by induction that $G'$ has an element $h'$ of order $p$.
Let $h$ be a preimage of $h'$ under the quotient epimorphism $\phi: G \to G'$.
Then:
- $\paren {h'}^p = e'$
where $e'$ is the identity of $G'$.
Thus $h^p \in \gen g$ so $\paren {h^p}^n = \paren {h^n}^p = e$.
Thus either $h^n$ has order $p$ or $h^n = e$.
If $h^n = e$ then $\paren {h'}^n = e'$.
But since $p$ is the order of $h'$ it would follow that $p \divides n$, contrary to assumption.
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Theorem $31$