# 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

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What does this notation $[S:1]$ mean? Is it the index of the trivial subgroup in $S$? – Damian Sobota Oct 26 '11 at 1:40
In general, $|ST| = |S||T|/|S\cap T|$. Moreover, $ST\neq\langle S,T\rangle$ in general. But you cannot conclude that $|\langle S,T\rangle|\geq |S||T|$, which is what you are claiming. For one thing take $S=T$. Then you are asserting that $|T|\geq |T|^2$, which would be pretty hard for most groups. – Arturo Magidin Oct 26 '11 at 1:55
@Damian Yes, you are correct. It is the index of the trivial subgroup in $S$. In other words, it is the number of elements of $S$. – Srivatsan Oct 26 '11 at 1:58

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 $s_0T \ne s_1T$?
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.
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