Center is Intersection of Centralizers

Theorem
The center of a group is the intersection of all the centralizers of the elements of that group:


 * $\displaystyle \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$

Proof
Denote $Z = \map Z G$ and $C = \displaystyle \bigcap_{g \mathop \in G} \map {C_G} g$ for simplicity.

By definition of set equality, it suffices to prove $Z \subseteq C$ and $C \subseteq Z$.

$Z$ is contained in $C$
Suppose that $x \in Z$.

Then from the definition of center:


 * $\forall g \in G: x g = g x$

By definition of centralizer, this corresponds to:


 * $\forall g \in G: x \in \map {C_G} g$

Therefore we have, by definition of set intersection:


 * $\displaystyle x \in \bigcap_{g \mathop \in G} \map {C_G} g = C$

Hence $x \in C$.

It follows by definition of subset that $Z \subseteq C$.

$C$ is contained in $Z$
Suppose now that $x \in C$.

Then, by definition of intersection:


 * $\forall g \in G: x \in \map {C_G} g$

That is, using the definition of centralizer:


 * $\forall g \in G: x g = g x$

By definition of the center, this means:


 * $x \in \map Z G = Z$

Hence $x \in Z$.

It follows that $C \subseteq Z$.

Therefore, we have established that:
 * $x \in Z \iff x \in C$

By definition of set equality:
 * $Z = C$