If $p<q$ are primes and $\mathrm{\mid }G\mathrm{\mid }=pq$, then $G$ has a normal subgroup of order $q$

Given: Primes $p<q$ and a group $G$ with $\mathrm{\mid }G\mathrm{\mid }=pq$.

If a prime divides the order of a finite group, the group has an element of that prime order (Cauchy's theorem: if a prime $p$ divides $\mathrm{\mid }G\mathrm{\mid }$, then $G$ has an element of order $p$).

The left-coset action of $G$ on $G\mathrm{/}Q$ is the homomorphism $\rho :G\to \mathrm{Sym}(G\mathrm{/}Q)$ with $\ker \rho ={\mathrm{Core}}_G(Q)$ (Left multiplication on $G\mathrm{/}H$ is transitive, has stabiliser $H$ at $H$, and has kernel ${\mathrm{Core}}_G(H)$).

The core ${\mathrm{Core}}_G(Q)$ is a normal subgroup of $G$ satisfying ${\mathrm{Core}}_G(Q)\le Q$ (${\mathrm{Core}}_G(H)$ is the largest normal subgroup of $G$ contained in $H$).

If $g$ has order $q$, the powers $g^0,\dots ,g^{q-1}$ are pairwise distinct and exhaust $⟨g⟩$, so $\mathrm{\mid }⟨g⟩\mathrm{\mid }=q$ (If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$).

Lagrange's theorem gives $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for a subgroup $H$ of a finite group (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

A set of $p$ elements has $p!$ bijections, where $p!=1\cdot 2\cdots p$ (A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality, The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$).

The image of a homomorphism is a subgroup and its kernel is normal; moreover the first isomorphism theorem identifies the quotient by the kernel with the image (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$).

For finite $G$ and normal $K$, $\mathrm{\mid }G\mathrm{/}K\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }K\mathrm{\mid }$ (If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$).

If a prime divides a product, it divides one of the factors (Euclid's lemma: if $p$ is prime and $p\mid ab$ then $p\mid a$ or $p\mid b$).