Cauchy's Group Theorem/Proof 1

Proof
Let $\order G$ be a prime number.

Then from Prime Group is Cyclic $G$ has a generator $\left\langle{g}\right\rangle$ 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 $\left\langle{g}\right\rangle$ has an element $g$ of order $p$.

Suppose $p \nmid n$.

From Subgroup of Abelian Group is Normal, $\left\langle{g}\right\rangle$ is normal in $G$.

Consider the quotient group $G' = \dfrac G {\left\langle{g}\right\rangle}$.

As $p \nmid n$, $\left\langle{e}\right\rangle \subsetneq \left\langle{g}\right\rangle \subsetneq G$.

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:
 * $\left({h'}\right)^p = e'$

where $e'$ is the identity of $G'$.

Thus $h^p \in \left\langle{g}\right\rangle$ so $\left({h^p}\right)^n = \left({h^n}\right)^p = e$.

Thus either $h^n$ has order $p$ or $h^n = e$.

If $h^n = e$ then $\left({h'}\right)^n = e'$.

But since $p$ is the order of $h'$ it would follow that $p \mathrel \backslash n$, contrary to assumption.