Let $G$ be a finite group, $H \lhd G$ If $[G:H]=m$ and $\lvert H \rvert=n$ with $gcd(m,n)=1$. Prove that $H$ is the only subgroup of G of order $n$. Let $G$ be a finite group and $H \lhd G$ Suppose that $[G:H]=m$ and $\lvert H \rvert=n$ with $gcd(m,n)=1$. Prove that $H$ is the only subgroup of G of order $n$. 
So, essentially, $\lvert G \rvert < \infty, H \lhd G, with [G:H]=m$
$\lvert H \rvert =n$, with gcd$(m,n)=1$ I have to show that $H$ is the only subgroup of order $n$
If I suppose $K < G, \lvert K \rvert =n$, I need to show $K=H$
$\lvert G \rvert =G$, if we say $\lvert H \rvert =3\,\, and\,\, \lvert K \rvert=3 \,\,\,\$Suppose $G \subset G, K 

So, $H \cap K < H \to \lvert H \cap K \rvert$ if $\lvert K \rvert =3$ and $\lvert H \cap K \rvert =1 \,\,or \,\,3$
If $\lvert HK \rvert = \frac{\lvert K \rvert \lvert H \rvert}{\lvert K \cap H \rvert}$ if $\lvert H \cap K \rvert =1$
I'm not sure how to go about some of this. 
 A: suppose $g$ is an element of $G$ whose order divides $n$. then if $\phi:G \to G/H$ let $g'=\phi(g)$ and $e,e'$ denote the identities in $G$ and $G/H$
then we have:
$$
g'^n = e'
$$
(by inheritance from $g^n=e$)
and
$$
g'^m = e'
$$
(since $|G/H|=m$)
$(m,n)=1$ means we may choose integers $p,q$ such that $pm+qn=1$ hence:
$$g' = g'^{pm+qn} =(g'^m)^p (g'^n)^q = e'^p e'^q = e'
$$
this shows any element whose order divides $n$ lies in $H$, a slightly stronger statement than the assertion to be proved
A: You can't assume any particular values for $m$ or $n$. The statement is to be proved for all values.
You are on the right track though. To begin, suppose $K$ is another subgroup of order $n$ as you do,  and then consider the size of $\phi(K)$, where $\phi:G\to G/H$ is the canonical projection map. What are a couple of things you can say about its size using $|K|=n$ and $|G/H|=m$?
A: The argument given by whacka's is less wordy, but you can also argue as follows:
Assume the hypotheses. Suppose $K$ is a subgroup of order $|H|$. Then $HK$ is a subgroup of $G$. We have $|HK| = \frac{|H||K|}{|H \cap K|} = \frac{n^{2}}{|H \cap K|}$. By LaGrange's $\frac{|HK|}{n} = \frac{n}{|H \cap K|}$ is an integer (because $H \cap K$ is a subgroup of both $K$ and $H$). Also, $|HK|$ divides $|G| = [G:H]|H| = mn$. It follows that $\frac{|HK|}{n}$ divides $m$. But since $m$ and $n$ are relatively prime, it follows that $m$ and $\frac{n}{|H \cap K|}$ are relatively prime. It follows that $1 = \frac{|HK|}{n} = \frac{n}{|H \cap K|} \implies |K| = n = |H\cap K|$. Since $H \cap K$ is a subgroup of $K$ with order $|K|$ it follows that $K$ is a subgroup of $H$, but since they have the same orders, $H = K$. 
A: As whacka said: you have that $\phi(K)$ is a subgroup of $G/H$. Show that the order of that group is divisor of $n$, but $gcd(m,n)=1$ which means that $|\phi(K)|=1$. Hence, $K$ is kernel of this cannonical projection, but by definition $H$ is kernel. Hence, $H$=$K$.
