Solved: Prove |HK|=|H||K| When H, K are Subgroups of G and H\capK = <e>

  • Thread starter chycachrrycol
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In summary, the notation "<e>" in this context represents the identity element in a group, which does not change the element when multiplied by any other element. The cardinality of HK is equal to the product of the cardinalities of H and K, as all possible combinations of elements from H and K are included in HK. An example is provided to illustrate the proof, where for a group G with subgroups H and K, |HK| = |H||K|. This proof is applicable to all groups as it relies on the fact that the intersection of H and K is equal to the identity element. This proof can also be extended to any number of subgroups, as long as the intersection of all subgroups is equal to the
  • #1
chycachrrycol
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So the problem is if H and K are subgroups of G with HK = {hk [tex]\in[/tex] G| h [tex]\in[/tex] H, k [tex]\in[/tex] K}. If we know that H[tex]\cap[/tex]K = <e>, show |HK|= |H||K|

My work so far:
h, j [tex]\in[/tex] H
k,l [tex]\in[/tex] K
i know that if hk = jl then j[tex]^{-1}[/tex]h = lk[tex]^{-1}[/tex]
But I'm not sure what to do from here.
 
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  • #2
j^(-1)h belongs to H, right? What about lk^(-1)?
 

FAQ: Solved: Prove |HK|=|H||K| When H, K are Subgroups of G and H\capK = <e>

1. What does the notation "" mean in this context?

The notation "" represents the identity element in a group, which is an element that when multiplied with any other element in the group, results in that same element. In other words, multiplying an element by the identity element does not change the element.

2. How is the cardinality of HK related to the cardinality of H and K?

The cardinality of HK, denoted as |HK|, is equal to the product of the cardinalities of H and K, denoted as |H| and |K| respectively. This is because HK consists of all possible combinations of elements from H and K, so the number of elements in HK is equal to the number of elements in H multiplied by the number of elements in K.

3. Can you provide an example to illustrate the proof?

Yes, let's consider a group G with subgroups H = {1, 2, 3} and K = {2, 4, 6}. The identity element in this group is 1. The intersection of H and K is the set {2}, which is equal to the identity element. Therefore, by the proof, we know that the cardinality of HK is equal to the cardinality of H multiplied by the cardinality of K. In this case, |H| = 3 and |K| = 3, so |HK| = 3 x 3 = 9. We can verify this by listing out all possible combinations of elements from H and K: {2, 4, 6, 8, 10, 12, 3, 6, 9}. The cardinality of this set is indeed 9.

4. Is this proof applicable to all groups?

Yes, this proof is applicable to all groups. This is because the proof relies on the fact that the intersection of H and K is equal to the identity element, which is a property of all groups. Therefore, the proof can be used to prove |HK| = |H||K| for any groups H and K.

5. Can this proof be extended to more than two subgroups?

Yes, this proof can be extended to any number of subgroups. The key is to show that the intersection of all the subgroups is equal to the identity element, and then use the same logic to prove that the cardinality of the product of all the subgroups is equal to the product of their individual cardinalities.

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