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

[PsychonautQQ]

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.

 

[lavinia]

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.

 

[PsychonautQQ]

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?

 

[lavinia]

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

 

[PsychonautQQ]

Wow thanks, that really helps. Can you prove why anything in the core is also in H?

 

[lavinia]

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.

 

[lavinia]

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.