Proving |HK| = |H||K|/|H∩K| in Group Theory

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In summary, the conversation discusses the proof that |HK|=\frac{|H||K|}{|H\cap K|} when G is a group and H and K are subgroups of G. The explanation involves a basic counting argument and demonstrates how elements in HK can be counted twice due to the presence of inverses in H and K. A hint is also provided for further understanding of the concept.
  • #1
rakalakalili
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Homework Statement


If G is a group, and H and K are subgroups of G, prove:
[tex]|HK|=\frac{|H||K|}{|H\cap K|}[/tex]

Homework Equations


The Attempt at a Solution


A basic counting argument can be made by saying that HK consists of an element of H with an element of K, so you have |H||K| elements. What I don't quite see is how this creates over counting by a factor of |H n k|. A hint would be much appreciated!
 
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  • #2
If h is in H and k is in K,

hk=hg g-1k

For any choice of g. In particular, if hg is in H and g-1k is in K, then we've counted hk twice.
 
  • #3
So if hg is in H, then that implies g is in H, which implies that g-1 is also in H. A similar argument shows that g and g-1 are in K. So for each element g of HK, hk will be equal to hgg^-1k, and that is where the extra count come in. Thank you!
 

FAQ: Proving |HK| = |H||K|/|H∩K| in Group Theory

1. What is the "Order of Operations" in Algebra?

The Order of Operations in Algebra is a set of rules that dictate the order in which mathematical operations should be performed in an expression. The acronym "PEMDAS" is often used to remember the order: Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

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