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

• PsychonautQQ
In summary: For instance, the identity element is in the intersection of all of the conjugates, but it is not in the core.Try to come up with and example where the core of H is greater that the identity element but not all of H.
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.

PsychonautQQ said:
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?

PsychonautQQ said:
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?

PsychonautQQ said:
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
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.

PsychonautQQ

## 1. Why is it important to know that the core of a subgroup is contained in the subgroup?

Understanding the relationship between the core of a subgroup and the subgroup itself is crucial in group theory, as it helps us to identify and classify different types of subgroups. It also allows us to better understand the structure and properties of the subgroup and its interactions with the larger group.

## 2. How is the core of a subgroup related to the normality of the subgroup?

The core of a subgroup is closely related to the concept of normality. In fact, a subgroup is normal if and only if its core is equal to the subgroup itself. This means that the core of a subgroup is a powerful tool in determining whether a subgroup is normal or not.

## 3. Can the core of a subgroup be larger than the subgroup itself?

No, the core of a subgroup cannot be larger than the subgroup. This is because the core is defined as the intersection of all conjugates of the subgroup, and by definition, a subgroup is contained in all of its conjugates.

## 4. How can we prove that the core of a subgroup is contained in the subgroup?

There are a few different ways to prove that the core of a subgroup is contained in the subgroup. One approach is to use the definition of the core as the intersection of conjugates and show that each conjugate is contained in the subgroup. Another approach is to use the normality of the subgroup and show that the core is equal to the subgroup.

## 5. Can the core of a subgroup be non-trivial?

Yes, the core of a subgroup can be non-trivial, meaning it is not just the identity element. This can happen when the subgroup is not normal, as the core will then be a proper subgroup contained in the subgroup. However, in the case of a normal subgroup, the core will always be trivial as it will be equal to the subgroup itself.

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