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I am stuck with the following problem. I am sure it cannot be that hard since it is intuitively true, but I can't find a way to prove it.

Let $A,B,C \leq G$ where $G$ is a finite group. Suppose moreover that $B \leq A$. Then $\vert A:B \vert \geq \vert C \cap A : C \cap B \vert$.

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it looks to me as a statement about integers. Have you tried to look at it that way? – Weltschmerz May 17 '11 at 12:18
@Weltschmerz: Not really but it looks worth it. On the other hand, I think Myself's answer is perfect. I had completely forgot the fact that $\vert A \vert \cdot \vert C \vert = \vert A \cap C \vert \cdot \vert AC \vert$. This proves instantly the implication. – Thomas Connor May 17 '11 at 12:26
up vote 5 down vote accepted

Asked is to show that $\frac{|A|}{|B|} \geq \frac{|A\cap C|}{|C\cap B|}$ since everything is finite.

We may use that $|A|\cdot |C| = |A\cap C|\cdot |AC|$ and similarly for $B$ and $C$, this amounts to showing that $|AC| \geq |BC|$, which is obvious since $B\subseteq A$.

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