If $H\unlhd G$ with $(|H|,[G:H])=1$ then $H$ is the unique such subgroup in $G$. Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman:

Let $G$ be a finite group, and let $H$ be a normal subgroup with $(|H|,[G:H])=1$. Prove that $H$ is the unique such subgroup in $G$.

I assumed there was another normal subgroup like $H$, say $K$, such that $(|K|,[G:K])=1$. My aim was to show that $[K: K\cap H]=1 $ that was not held if I didn’t suppose $|H|=|K|$ . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks.
 A: Let $L$ be any subgroup of $G$ with order $|H|$. Let $p$ be the natural projection $G \to \frac{G}{H}$. Then $p(L)$ is a quotient group of $L$, so $|p(L)|$ divides $|L|=|H|$. But $p(L)$ is also a subgroup of $\frac{G}{H}$, so $|p(L)|$ also divides $|G:H|$. By the coprimality hypothesis, $|p(L)|=1$ so $p$ is trivial on $L$. This means that $L \subseteq H$. Finally $L=H$ because those two subgroups have the same cardinality.
A: One way to see this is to look at the quotient map $f\colon G \to G/H$. Using Lagrange's theorem a couple of times, what can you say about the order of $f(K)$? [This isn't far from proving a well known formula for $\#(HK)$.]
A: Consider the more general result proven by user Gastón Burrull here. It states

If $K\triangleleft G$, $|K|$ finite, $H\leq G$, $[G:H]$ finite and $|K|$, $[G:H]$  are relatively prime then $H\leq K$

In other words, if the index of a subgroup is coprime to the cardinality of a normal subgroup, then it is contained in that normal subgroup. So if $K$ is any other subgroup of cardinality $|H|$, then since the group is finite, $H$ and $K$ have the same index, hence the index of $K$ is coprime to $|H|$, and so $K\leq H$. But since $K$ and $H$ are finite subgroups of equal order, $K=H$. So $H$ is the only subgroup of order $|H|$.
A: To make things explicit: Assume $\lvert H \rvert = \lvert K \rvert$. 
As a consequence of the 2nd isomorphism theorem, $HK \le G$ and 
$$\lvert HK \rvert 
= \frac{\lvert H \rvert \lvert K \rvert}{\lvert H \cap K \rvert} 
= \frac{\lvert H \rvert^2}{\lvert H \cap K \rvert}$$
By Lagrange's Theorem:
$$\lvert G \rvert = [G:H] \lvert H \rvert 
= [G:HK] \lvert HK \rvert $$
We derive 
$$[G:H] = [G:HK] \frac{\lvert H \rvert}{\lvert H \cap K \rvert}$$
If $\lvert H \cap K \rvert < \lvert H \rvert$ then there is some factor $n \ne 1$ of $\lvert H \rvert$ on the RHS (note: $\lvert H \vert /\lvert H \cap K \rvert 
 = n \in \mathbb N$). Thus $n \mid [G:H] $, a contradiction that $\operatorname{gcd}(\lvert H \rvert, [G:H]) = 1$. 
We conclude $\lvert H \cap K \rvert = \lvert H \rvert$ and since $\lvert H \rvert = \lvert K \rvert$ we can conclude $H = K$. 
A: Alternatively, you might note that if $K$ is any other subgroup of order $|H|$, whether or not $K$ is normal (but assuming $H$ is normal), then $HK$ is a subgroup of $G$, so its order must divide $|G|$.
A: This machinery is somewhat high powered, but it should explain what the book is really trying to get at.  
Hall subgroups are generalizations of Sylow subgroups for multiple primes.  If we denote by $\nu_G(p)$ the largest power of $p$ dividing $|G|$, we see that Sylow $p$-subgroups are those subgroups of order $\nu_G(p)$.  Hall $\pi$-subgroups, where $\pi$ is a set of prime numbers, are subgroups of order $\prod_{p\in \Pi} \nu_G(p)$.  Alternatively, a Hall $\pi$-subgroup $H$ is a subgroup of order divisible by each $p\in \Pi$ such that $[G:H]$ is coprime to $|H|$.  Note that Hall subgroups of a group $G$ may not always exist for each $\pi\subset\{p\in \mathbb{P}|p\mid |G|\}$ - actually, this holds if and only if $G$ is solvable. 
As it turns out, a lot of the stuff that works for Sylow subgroups works for Hall subgroups.  In particular, when Hall subgroups exist, they are all conjugate, so much like how normal Sylow subgroups must be unique, so must normal Hall subgroups.  This is part of Hall's theorem, which I will not prove here (but it follows fairly easily by inducting on the size of a minimal normal subgroup).
