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This question/thread is continuation from here Also I.N Herstein in Page 40 Question 6 has brought up the same question

My approach: We know the property $\left | G:K \right |\leq \left | G:H \right|\left | H:K \right | $ if $K$ is the subgroup of $H$ which is the subgroup of $G$

So if we can prove $H\cap K$ is a subgroup of H, then the work is done

We know, $H \cap K \subseteq H$

$\Rightarrow$ if $h_1,h_2 \in H\cap K$, then $h_1,h_2 \in H$ also if $h_1,h_2 \in H\cap K$, then $h_1,h_2 \in K$

$\Rightarrow$ $h_1\cdot h_2 \in H$ and $h_1\cdot h_2 \in K$

$\Rightarrow$ $h_1\cdot h_2 \in H \cap K$

Thus its closed under product

Similarly, we can argue about $1_G \in H \cap K $ [since $H,K$ are both subgroups and thus identity must be common to both of them]

$\Rightarrow$ inverse exists in $H \cap K$

Thus $H \cap K$ is a subgroup of $H$ as well as $K$

Hence applying the property

$\left | G:K \right | \leq \left | G:H \right|\left | H:K \right | $

We get if $\left | G:H \right|$ = $m$ [since finite] and $\left | H:H \cap K \right|$=$k$

$\Rightarrow \left | G:K \right | \leq mk$

Equality holds when $G$ is a finite group, $mk$ being the upper bound of the index of $H\cap K$ in $G$

Soham

Please do raise any questions if I have gone wrong somewhere

Soham

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    @GeoffRobinson Yes, I agree.2012-08-01

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In light of Geoff comment, an approach came to me. I hope it is right:

If $(H∩K)x$ be a right coset of subgroup $H∩K$ in $G$ then we have $(H∩K)x=Hx∩Kx$. So every right coset of $H∩K$ is an intersection of a right coset of $H$ and a right coset of $K$. In other words every right coset of $H∩K$ is as $Ha∩Kb$ where $a$ and $b$ in $G$ . This means that the number of distinct right cosets of $H∩K$ in $G$ is less or equal than such these combinations. Or $[G:H∩K]≤[G:H][G:K]$.

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    Your point of view is pretty darn upper to me! :-)2013-03-12