Prove that $lcm(m,n) \leq |G : H \cap K| \leq mn$ where $|G:H| = m$ and $|G:k | = n$ I am trying to show that $lcm(m,n) \leq |G : H \cap K| \leq mn$ where $|G:H| = m < \infty$ and $|G:k | = n < \infty$. Particularly I am trying to solve it using group actions, kernels and stabilizers (perhaps not necessarily using the orbit stabilizer theorem). However, I feel completely lost on how to start approaching this problem. Anyone has any hints or suggestions on how to solve this problem? Thanks! 
 A: Write $G/H$ for the set of cosets $xH$ for $x\in G$ (note this is not
necessarily a group!). Then $G$ acts on the left on $G/H\times G/K$
by $g(xH,yK)=(gxH,gyK)$. The stabiliser of $(H,K)$ is $H\cap K$.
Now what is the size of the orbit of $(H,K)$ and how does that compare
to the size of $G/H\times G/K$?
For the first inequality, can you show that $|G:H|$ is a  factor
of $|G:H\cap K|$?
A: Hints. Let $A=\{gH: g\in G\}$ and $B=\{gK: g\in G\}$ and $C=\{g\cdot (H\cap K):g\in G\}.$  
(1)...   Show that $C=\{a\cap b: a\in A\land b\in B\}$ \ $\{\phi\}$ by showing that if $a\in A$ and $b\in B$ then  $ a\cap b \ne \phi$ iff  there exists $x\in G$ such that $(xH)\cap (xK)=x(H\cap K)=a\cap b. $ Hence $|C|\leq |A|\cdot |B|=mn.$
(2)...   For $a\in A$ let $a^*=\{a\cap b: b\in B\}$ \ $\{\phi\}.$ Use (1) to show that if $a_1=g_1H\in A$ and  $a_2=g_2H\in  A$ then $F(X)=g_2g_1^{-1}X $ is a bijection from $a_1^*$ to $a_2^*$, and since $a_1^*\ne \phi$, we obtain the result that $m$ divides $|\cup \{a^* :a\in A\}|=|C|$. Interchanging the roles of $H$ and $K,$ we also find that $n$ divides $|C|$.
None of this is deep. Just use the def'ns and the very basic properties of groups.
