Why is the core of a subgroup contained in the subgroup?

  • #1
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Let H be a subgroup of G, then:
Core H = {a in G | a is an element of gHg^(-1) for all g in G} = The intersection of all conjugates of H in G

My book goes on to say that every element of Core H is in H itself because H is a conjugate to itself. Previously, I understood that H was a conjugate to itself because 1H1^(-1) = H, where 1 is the identity. However, the definition of Core H is the *intersection* of all conjugates, and it seems that core of H would be included in H only if gHg^(-1) = H for all g in G, (aka H is normal).

What am I not understanding here? Anyone who can shed any light what so ever is greatly appreciated.
 

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  • #2
lavinia
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Let H be a subgroup of G, then:
Core H = {a in G | a is an element of gHg^(-1) for all g in G} = The intersection of all conjugates of H in G

My book goes on to say that every element of Core H is in H itself because H is a conjugate to itself. Previously, I understood that H was a conjugate to itself because 1H1^(-1) = H, where 1 is the identity. However, the definition of Core H is the *intersection* of all conjugates, and it seems that core of H would be included in H only if gHg^(-1) = H for all g in G, (aka H is normal).

What am I not understanding here? Anyone who can shed any light what so ever is greatly appreciated.
The intersection of all of the conjugates must be in H since H is one of the conjugates. For instance, the identity element is in the intersection of all of the conjugates.
 
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  • #3
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I understand the identity element is in the intersection of all of the conjugates, but how would you prove that an arbitrary h in H is in the intersection?
 
  • #4
lavinia
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I understand the identity element is in the intersection of all of the conjugates, but how would you prove that an arbitrary h in H is in the intersection?
Not all of the elements of H are in the intersection - in general.

For instance take the group with two generators a and b with the relation
## a^2 = b^4 = 1##
## aba = b^3##

##a## generates a subgroup,##H##, of order 2.

##H = {a,1}##

## bHb^{-1} = bab^{-1}, 1##

but
## bab^{-1} = bab^{3} = ab^{2}##

So a is not in the core of H
 
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  • #5
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Wow thanks, that really helps. Can you prove why anything in the core is also in H?
 
  • #6
lavinia
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Wow thanks, that really helps. Can you prove why anything in the core is also in H?
I misread your question. Below I am explaining why not all of H need be in the core.

Well the identity always is. But after that there are no guarantees. If H is a normal subgroup then all of H is in the core. In the example I gave - which BTW is the dihedral group of order 8 - I purposefully chose a not normal subgroup.

Try to come up with and example where the core of H is greater that the identity element but not all of H.
 
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  • #7
lavinia
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Sorry I misread your question. I was saying that not all of H needs to be in the core. The reason that the core must be contained in H is that His one of the sets in the intersection. So if isn't in H it is not in the core.
 
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