How to prove $|S||T|\leq |S \cap T ||\langle S, T\rangle|$? If $S$ and $T$ are subgroups of a finite group $G$, prove that $|S||T|\leq |S \cap T||\langle S, T\rangle|$.
My approach is: since $\langle S, T\rangle$ is the smallest subgroup containg $S$ and $T$, it should at least contain all the elements of the form $st$ where $s$ belongs to $S$ and $t$ belongs to $T$.
Since $S$ has $|S|$ elements, $T$ has $|T|$ elements, $|\langle S, T\rangle| \geq |S||T|$.
Since $S$ and $T$ are both subgroups of $G$, they should at least both contain $1$ (unit). Therefore $|S \cap T|\geq 1$.
Putting them together, $|S||T|\leq |S \cap T||\langle S, T\rangle|$.
Is my proof right? Thanks
 A: I don't think this quite works, because you cannot assume that those elements of form $st$ are all unique.  In particular, if $S$ is a subgroup of $T$, then $\langle S,T\rangle$ is simply $T$, so the inequality $[\langle S,T\rangle: 1] \geq [S:1][T:1]$ does not hold (unless $S$ is the trivial subgroup).
Do you have any results regarding the join and intersection that you can use?  For example, your problem can be rearranged a bit to $[\langle S,T \rangle:T]\geq[S:S\cap T]$, which is a result I believe I've seen in texts before.
A: As AMPerrine noted, your problem is equivalent to proving that $[\langle S,T\rangle]\ge[S:S\cap T]$. To prove this, it suffices to show that if $s_0,s_1\in S$, and the cosets $s_0(S\cap T)$ and $s_1(S\cap T)$ are distinct, then the cosets $s_0T$ and $s_1T$ are also distinct. (Why?)
Suppose that $s_0,s_1\in S$, and $s_0(S\cap T)\ne s_1(S\cap T)$. Then there must be some $t\in S\cap T$ such that $s_0t\notin s_1(S\cap T)$, or, equivalently, such that $s_0ts_1^{-1}\notin S\cap T$. Since $t\in S$, we know that $s_0 t s_1^{-1}\in S$; thus, we must have $s_0ts_1^{-1}\notin T$. Can you see how to finish this off to show that $s_0T \ne s_1T$?
