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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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thanks @rschwieb I missed that –  Soham Aug 1 '12 at 12:08
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How do you know that $H\cap K$ is finite index in $H$? –  Grumpy Parsnip Aug 1 '12 at 12:13
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I would have thought Herstein would have taken it as a given that the intersection of two subgroups is a subgroup. The content of the question is probably to show the inequality you assume as given, (I can't remember whether that is theorem earlier in the text) and then, as Jim Conant points out, you need to justify why $[H:H \cap K]$ is finite. –  Geoff Robinson Aug 1 '12 at 12:18
    
Okay, thanks I think I have missed that part, further I was rightly thinking that Herstein surely wouldnt have marked it as starred to show intersection of two subgroups is a sg. :) Thanks let me work on it. –  Soham Aug 1 '12 at 12:30
    
If $H$ and $K$ be assumed to be finite subgroups pf $G$ then you can show that $[H:H∩K]≤[G:K]$. –  B. S. Aug 1 '12 at 12:44
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1 Answer

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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Looks good, but you lost me at the inequality. Why the number of right cosets of $H \cap K $ will affect the index ? –  Soham Aug 1 '12 at 13:12
    
And $"="$ can't be happen unless $([G:H],[G:K])=1$ –  B. S. Aug 1 '12 at 13:25
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@Babak: It is even true that $[H:H \cap K] \leq [G:K].$ The comment about when "=" holds is not correct. –  Geoff Robinson Aug 1 '12 at 13:38
    
@GeoffRobinson: For what I commented above, please have a look at my proof here. $[G:H]=m, [G:K]=n$ so since $[G:H∩K]=l$ is finite then as above $l≤mn$. But $[G:H∩K]=[G:H][H:H∩K]=[G:K][K:H∩K]$ and then $n|l,m|l$. Doesn't this lead us to $'='$? Thanks –  B. S. Aug 1 '12 at 16:48
    
If the indices are coprime, then you get equality. But you can have equality without coprime indices. Conside G a Klein 4-group, and $H,K$ of order $2$ (different subgroups). –  Geoff Robinson Aug 1 '12 at 18:46
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