Center is Intersection of Centralizers

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Theorem

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

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


Proof

Denote $Z = \map Z G$ and $C = \ds \bigcap_{g \mathop \in G} \map {C_G} g$ 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:

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

Hence $x \in C$.

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

$\Box$


$C$ 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 $C \subseteq Z$.

$\Box$


Therefore, we have established that:

$x \in Z \iff x \in C$

By definition of set equality:

$Z = C$

$\blacksquare$


Sources

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