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Suppose $H$ and $K$ are subgroups of a finite group $G$. Then I read $$ |HxK|=|H|[K:x^{-1}Hx\cap K] $$

where $x\in G$. How can this equality be derived?

I wanted to prove the equivalent equality $|HxK|/|H|=[K:x^{-1}Hx\cap K]$ by exhibiting a bijection of the cosets of $x^{-1}Hx\cap K$ in $K$ with cosets of $H$ in $HxK$, but I realize $HxK$ need not be a group, nor $H$ a subgroup. Thanks.

Source: This is the third statement of Exercise 5 of Jacobson's Basic Algebra I, page 53.

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But still $HxK$ is a union of right cosets of $H$, so if you can exhibit that bijection, you'll be fine. –  Jyrki Lahtonen May 26 '12 at 9:23
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See if this link help you usna.edu/Users/math/wdj/tonybook/gpthry/node44.html –  Babak S. May 26 '12 at 9:30
    
@JyrkiLahtonen Oh right, thanks! And the bijection is $(x^{-1}Hx\cap K)k\mapsto Hxk$. –  Adelaide Dokras May 26 '12 at 16:38
    
Thanks for the link @BabakSorouh, there is lots of good info there. –  Adelaide Dokras May 26 '12 at 16:38

1 Answer 1

up vote 4 down vote accepted

$K$ operates on the left cosets of $H$ by translation: $k(xH) := kxH$. Let $m := |\{kxH : k\in K\}|$ denote the size of the orbit of $xH$ under this operation. Then $KxH$ is the disjoint union of $m$ sets of the form $kxH$, $k\in K$, each of which has size $|H|$, so $|KxH| = m|H|$.

On the other hand, we know that the size of the orbit of $xH$ equals the index of the stabilizer. An element $k \in K$ is in the stabilizer iff $$kxH = xH \Leftrightarrow x^{-1}kx \in H \Leftrightarrow k \in xHx^{-1},$$ so the stabilizer is $K \cap xHx^{-1}$. So we have $$m = [K : K\cap xHx^{-1}] = [K : x^{-1}Kx \cap H]$$ (note that $K\cap xHx^{-1}$ and $x^{-1}Kx \cap H$ are conjugated subgroups of $K$, hence have the same index). Finally, $$|KxH| = m|H| = |H| [K : x^{-1}Kx \cap H].$$

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Thanks! ${}{}{}$ –  Adelaide Dokras May 26 '12 at 16:39

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