Proving that K is the unique subgroup of G having order |K|

Question

Let $G$ be a finite group with $K$ being a normal subgroup of $G$. If $(|K|, [G : K]) = 1$, prove that $K$ is the unique subgroup of $G$ having order $|K|$.

Answer 1

Consider the quotient map $p: G\rightarrow G/K$ and let $L$ be a subgroup which has the same order as $K$, the order of $p(L)$ divides $|K|$ (since it is the quotient of the order of $L$ and the order of the restriction of $p$ t $L$) and $|G/L|=[G:K]$ (since $p(L)$ is a subgroup of $G/K$), since $gcd(|K|, [G:K])=1$ we deduce that $|p(L)|=1$ and $L\subset K$, $L=K$ since $L$ and $K$ have the same cardinality.