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This question already has an answer here:

I want to prove following statement

If $H$ and $K$ are finite supgroups of a group $G$, then $|HK| = \frac{|H||K|}{H\cap K}$

This is a proposition 13 in Dummit's abstract algebrac chapter 3.3.

Intuitively, i can guess the order of $|HK|$ is proportial to $|H|$ and $|K|$ modulo some intersection.

The textbook states, \begin{align} HK = \cup_{h\in H} hK \end{align} then it suffices to show that there are $\frac{|H|}{|H\cap K|}$ distinct coset in this union.

I can understand the division of $HK$ as a sum of $hK$, but having trouble understanding the next step.

Can you explain this in more detail?

Perhaps other prove of this will be helpful.


I have found same problem in

How to prove that $|HK| = \dfrac{|H| \; |K|}{|H \cap K|}$?

Now i got some sense

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marked as duplicate by phy_math, Dietrich Burde abstract-algebra Apr 15 '17 at 8:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For $h,h' \in H$ you have $$hK = h'K \iff hh'^{-1} \in K \iff hh'^{-1}\in H\cap K$$ Therefore for fixed $h$ there are exactly $|H\cap K|$ elements $h'\in H$ such that $hK = h'K$, and $hK$ is of cardinal $|K|$. We also proved that the intersections of the cosets is empty.

Thus the result follows.

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