Center of Group is Subgroup

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Let $G$ be a group.

The center $\map Z G$ of $G$ is a subgroup of $G$.

Proof 1

For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).

Apply the Two-Step Subgroup Test:

Condition $(1)$

By the definition of identity, $e g = g e = g$ for all $g \in G$.

So, $e \in \map Z G$, meaning $\map Z G$ is non-empty.


Condition $(2)$

Suppose $a, b \in \map Z G$.

Using the associative property and the definition of center, we have:

$\forall g \in G: \paren {a b} g = a \paren {b g} = a \paren {g b} = \paren {a g} b = \paren {g a} b = g \paren {a b}$

Thus, $a b \in \map Z G$.


Condition $(3)$

Suppose $c \in \map Z G$. Then:

\(\ds c\) \(\in\) \(\ds \map Z G\)
\(\ds \leadsto \ \ \) \(\ds \forall g \in G: \, \) \(\ds c g\) \(=\) \(\ds g c\) Definition of Center of Group
\(\ds \leadsto \ \ \) \(\ds c^{-1} \paren {c g} c^{-1}\) \(=\) \(\ds c^{-1} \paren {g c} c^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {c^{-1} c} g c^{-1}\) \(=\) \(\ds c^{-1} g \paren {c c^{-1} }\)
\(\ds \leadsto \ \ \) \(\ds e g c^{-1}\) \(=\) \(\ds c^{-1} g e\)
\(\ds \leadsto \ \ \) \(\ds g c^{-1}\) \(=\) \(\ds c^{-1} g\)
\(\ds \leadsto \ \ \) \(\ds c^{-1}\) \(\in\) \(\ds \map Z G\)



$\map Z G \le G$


Proof 2

We have the result Center is Intersection of Centralizers.

That is, $\map Z G$ is the intersection of all the centralizers of $G$.

All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup.

Thus from Intersection of Subgroups is Subgroup, $\map Z G$ is also a subgroup of $G$.