Proving two subgroups with same cardinality are identical if one is normal I need to prove the following statement:
Let $G$ be a finite group, $H, K \subset G$ two subgroups, $gcd(Card(H), Card(G/H))=1, Card(H)=Card(K), H$ normal in G. Then $H=K.$
We know that Card($G$) = Card($H$) [G:H], so Card($G$) = Card($H$) Card($G/H$). According to one of the isomorphism theorems, $KH<G, (K\cap H) \triangleleft H $, and $f: K \rightarrow G/H$ is a group homomorphism with ker $ f= K\cap H$ and $K/(K \cap H) \cong KH/H.$
Somehow I can not connect  the dots. Can somebody help me.
Many thanks.
 A: Let $k \in K$ and consider $kH \in G/H$.  Denote $n = |K| = |H|$ and $|k| = m$.  Then since the order of $k$ divides the order of $K$, we have $m \mid n$.  
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$.  Since $kH \in G/H$, we also have that $r$ divides $|G/H|$.  If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$.  But this is a contradiction since we have assumed $\gcd(n, |G/H|) = 1$.
So it must  be that the order of $kH$ is $1$, which then implies that $k \in H$.  Since $k \in K$ was arbitrary we have that $K \leq H$.  But this implies that $H = K$ since they have the same order.
A: Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $\phi : G\to G/H$ be the quotient map. Consider $\phi(K)$. By the first isomorphism theorem, we have $\phi(K)\cong K/K\cap H$, so $|\phi(K)|=|K/K\cap H|$, and $|\phi(K)|$ divides $|K|=|H|$, however $\phi(K)$ is also a subgroup of $G/H$, so $|\phi(K)|$ divides $|G/H|$. Hence $|\phi(K)|$ divides $\operatorname{gcd}(|H|,|G/H|)=1$. Thus $|\phi(K)|=1$, so $K\subseteq H$, and since $|K|=|H|$, this implies $K=H$.
