Center is Intersection of Centralizers

Theorem

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

$\displaystyle Z \left({G}\right) = \bigcap_{g \in G} C_G \left({g}\right)$

Proof

Denote $Z = Z \left({G}\right)$ and $C = \displaystyle \bigcap_{g \in G} C_G \left({g}\right)$ for simplicity.

From Equality of Sets, 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:

$\forall g \in G: x \in C_G \left({g}\right)$

Therefore we have, by definition of set intersection:

$\displaystyle x \in \bigcap_{g \in G} C_G \left({g}\right) = C$

Hence $x \in C$. It follows that $Z \subseteq C$.

$\Box$

$C$ is contained in $Z$

Suppose now that $x \in C$.

Then, by definition of intersection:

$\forall g \in G: x \in C_G \left({g}\right)$

That is, using the definition of centralizer:

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

By definition of the center, this means:

$x \in Z \left({G}\right) = Z$

Hence $x \in Z$. It follows that $C \subseteq Z$.

$\Box$

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

From the definition of set equality, it follows that $Z = C$.

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 37$