$N$ is the only subgroup of order $|N|$ 
Let $G$ be a group of finite order. Let $N$ be a normal subgroup of $G$ and $\gcd(|N|,|G:N|)=1$. Show that $N$ is the only subgroup of order $|N|$.

Here is my attempt at a solution:
Suppose there is another subgroup $H$ of $G$ such that $|N|=|H|$ and $H\neq N$.
Since $N$ is normal then $HN$ is a subgroup of $G$.
By the Diamond Isomorphism Theorem:
$HN/N$ is isomorphic to $H/H\cap N$ $\implies$ $|HN/N|=|H/H\cap N|$.
$\frac{|HN|}{|N|} = \frac{|H|}{|H\cap N|}$ since the order are integers I can manipulate the equation to have:
$|HN||H\cap N|=|H||N|\implies |HN||H\cap N|=|N|^{2}\implies|HN|=|H\cap N|=|N| $.
But since $N\neq H$ we can't have $|HN|=|H\cap N|=|N|$ a contradiction.
Thus $H=N$.
Is my reasoning correct? I'm kind of doubtful because I have not even used the fact that $\gcd(|N|,|G:N|)=1$.
Thank you!
 A: I think your proof is correct except that I'm not sure I follow the following implication in your working:
$\left|HN\right|\left|H\cap N\right|=\left|N\right|^2$
$\implies \left|HN\right|=\left|H\cap N\right|=\left|N\right|$
without further justification/use of hypothesis. The implication seems to be based on the claim that there is exactly one factorization of the positive integer $\left|N\right|^2$ as the product of two positive integers; namely, $\left|N\right|^2=\left|N\right|\left|N\right|$. However, this claim is certainly false in general (e.g, $4^2=16=8\cdot 2$.). 
However, the step is salvageable if you use the hypothesis that $\left|N\right|$ and $\left|G:N\right|$ are coprime. Indeed, you can rewrite $\left|HN\right|\left|H\cap N\right|=\left|N\right|^2$ as:
$\frac{\left|HN\right|}{\left|N\right|}=\frac{\left|N\right|}{\left|H\cap N\right|}$.
Now, can you see how to use the hypothesis to finish off the proof? I am happy to provide further hints if you'd like.
I hope this helps!
A: Let $H \leq G$ with $|H|=|N|$. Then $H \to G \to G/N$ is trivial, because the order of its image is a divisor of $|H|=|N|$, but also of $|G/N|$. Hence $H \subseteq N$. Since they have the same order, $H=N$.
A: Perhaps you could avoid arguing by contradiction. And actually with no more effort you can show that any subgroup $H$ of order dividing $\lvert N \rvert$ is contained in $N$.
To prove this, the arguments are those that have already appeared. You compute $\lvert H N : N \rvert$ in two ways.
First, by Lagrange, it divides $\lvert G : N \rvert$. Explicitly,
$$
\lvert G \rvert = \lvert G : HN \rvert \cdot \lvert HN \rvert
=
\lvert G : HN \rvert \cdot \lvert HN : N \rvert \cdot \lvert N \rvert,
$$
and since $\lvert G \rvert =  \lvert G : N\rvert \cdot \lvert N \rvert$, we have
$$
\lvert G : N \rvert
=
\lvert G : HN \rvert \cdot \lvert HN : N \rvert \cdot.
$$
And then $\lvert H N : N \rvert = \lvert H : H \cap N \rvert$ divides the order of $H$, and thus the order of $\lvert N \rvert$.
