Show that centre of group is equal to intersection of centralizer

To show that the **center** of a group \( G \), denoted as \( Z(G) \), is equal to the **intersection of centralizers** of all elements in \( G \), follow these steps:

### Definitions:
1. **Center of a group**:
   The center of a group \( G \), denoted \( Z(G) \), is the set of elements in \( G \) that commute with every element of \( G \). Formally:
   \[
   Z(G) = \{ z \in G \mid zg = gz \ \text{for all} \ g \in G \}
   \]
   So, any element in the center commutes with every element in \( G \).

2. **Centralizer of an element**:
   The centralizer of an element \( g \in G \), denoted \( C_G(g) \), is the set of elements in \( G \) that commute with \( g \). Formally:
   \[
   C_G(g) = \{ x \in G \mid xg = gx \}
   \]
   So, any element in the centralizer of \( g \) commutes with \( g \).

### Goal:
We need to show:
\[
Z(G) = \bigcap_{g \in G} C_G(g)
\]
This means that the center of the group is the intersection of the centralizers of all elements in \( G \).

### Proof:

1. **Step 1: Show that \( Z(G) \subseteq \bigcap_{g \in G} C_G(g) \)**

   Let \( z \in Z(G) \). By the definition of the center, \( z \) commutes with every element of \( G \), i.e., for all \( g \in G \), we have:
   \[
   zg = gz
   \]
   But this implies that \( z \in C_G(g) \) for all \( g \in G \) because it commutes with every element \( g \). Therefore, \( z \) is in the intersection of all centralizers:
   \[
   z \in \bigcap_{g \in G} C_G(g)
   \]
   This shows that \( Z(G) \subseteq \bigcap_{g \in G} C_G(g) \).

2. **Step 2: Show that \( \bigcap_{g \in G} C_G(g) \subseteq Z(G) \)**

   Let \( x \in \bigcap_{g \in G} C_G(g) \). This means that for every \( g \in G \), \( x \in C_G(g) \), i.e.,
   \[
   xg = gx \quad \text{for all} \ g \in G
   \]
   Hence, \( x \) commutes with every element of \( G \), and by the definition of the center, this implies that \( x \in Z(G) \).

Thus, \( \bigcap_{g \in G} C_G(g) \subseteq Z(G) \).

### Conclusion:
Since both inclusions are true, we conclude that:
\[
Z(G) = \bigcap_{g \in G} C_G(g)
\]
This completes the proof.

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Do you have any questions or would you like more details on any part of the proof? 

Here are 5 questions for further exploration:
1. What are some examples of groups where the center is non-trivial?
2. How is the centralizer of an element related to the conjugacy class of that element?
3. What is the relationship between the center and the normalizer of a subgroup?
4. Can the center of a group be trivial? Give examples.
5. How do centralizers behave in non-abelian groups?

**Tip:** In finite groups, the order of the center always divides the order of the group, a property that can help in understanding group structure.

Show that the centre of group is equal to intersection of centralizer.

Explore the proof that the center of a group is the intersection of centralizers of all elements. This problem delves into group theory concepts such as the center of a group and centralizers, offering a step-by-step explanation suitable for university-level students or advanced high school learners.

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