Show that G = HK (Subgroups) Let $H$ and $K$ be subgroups of $G$ and  $gcd(|G:H|,|G:K|)=1$.
Show that $G=HK$
I would appreciate a hint. 
Using Lagrange's Theorem we get
$|G|= |G:H||H|$ and
$|G|= |G:K||K|$
But that's not getting me there. 
 A: We are not given that $G$ is finite, though the indices of $H$ and $K$ must be finite in order to talk about their greatest common divisor. That means that $H$ contains a normal subgroup $N_1$ of finite index in $G$, and $K$ contains a normal subgroup $N_2$ of finite index in $G$, and hence $N_1\cap N_2$ is of finite index in $G$, so by moding out by this subgroup, we could pass to the finite case. But we do not need to. 
However, contra Nick's answer, we do not know that $HK$ is a subgroup ahead of time, so we cannot argue via Lagrange's Theorem with $HK$ unless and until we know it is a subgroup. We also cannot "divide by $H$" unless we do pass to the finite case as above. 
Lemma. $[H:H\cap K]\leq [G:K]$, with equality if and only if $G=HK$.
Proof. Define a map from the cosets of $H\cap K$ in $H$ to the cosets of $K$ in $G$ by $f(x(H\cap K)) = xK$. 
This is well defined: if $x(H\cap K) = y(H\cap K)$, then $y^{-1}x\in H\cap K$, hence $xK = yK$. It is also one-to-one: if $xK=yK$ with $x,y\in H$, then $y^{-1}x\in H\cap K$, so $x(H\cap K)=y(H\cap K)$. This gives the inequality.
Now, if $HK=G$, then given $g\in G$, we can write $g=hk$ with $h\in H$ and $k\in K$; and then $gK = hK = f(h(H\cap K))$, so $f$ is also onto. Conversely, if $f$ is onto and $g\in G$, then there exists $h\in H$ such that $gK = f(h(H\cap K)) = hK$, hence there exists $k\in K$ such that $g = hk$. Thus, $G=HK$ as claimed.~$\Box$
Similarly, $[K:H\cap K]\leq [G:H]$ with equality if and only if $G=HK$.
Note in particular that since $[G:H]$ and $[G:K]$ are both finite, it follows that $[H:H\cap K]$ and $[K:K\cap H]$ are both finite, and so $K\cap H$ has finite index in $G$. 
Now, $[G:H\cap K] = [G:H][H:H\cap K]$ and $[G:H\cap K] = [G:K][K:H\cap K]$. Thus, $[G:H]$ divides $[G:K][K:H\cap K]$, and since $\gcd([G:H],[G:K])=1$, then $[G:H]$ divides $[K:H\cap K]$. Symmetrically, $[G:K]$ divides $[H:H\cap K]$. 
But $[H:H\cap K]\leq [G:K]$, and so from the fact that $[G:K]$ divides $[H:H\cap K]$, we conclude that $[G:K]=[H:H\cap K]$, and hence that $G=HK$ by the Lemma, as claimed. 
