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

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
Gold Member
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
Gold Member
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
Gold Member
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.

Last edited:
• PsychonautQQ
lavinia
• 