Center of Group is Subgroup
Theorem
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.
$\Box$
Condition $(2)$
Suppose $a, b \in \map Z G$.
Using the associative property and the definition of center, we have:
\(\ds \forall g \in G: \, \) | \(\ds \paren {a b} g\) | \(=\) | \(\ds a \paren {b g}\) | Definition of Associative Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {g b}\) | Definition of Center of Group | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a g} b\) | Definition of Associative Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {g a} b\) | Definition of Center of Group | |||||||||||
\(\ds \) | \(=\) | \(\ds g \paren {a b}\) | Definition of Associative Operation |
Thus, $a b \in \map Z G$.
$\Box$
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 | |||||||||
\(\ds \leadsto \ \ \) | \(\ds c^{-1} \paren {c g} c^{-1}\) | \(=\) | \(\ds c^{-1} \paren {g c} c^{-1}\) | Group Axiom $\text G 0$: Closure | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {c^{-1} c} g c^{-1}\) | \(=\) | \(\ds c^{-1} g \paren {c c^{-1} }\) | Definition of Associative Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e g c^{-1}\) | \(=\) | \(\ds c^{-1} g e\) | Definition of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds g c^{-1}\) | \(=\) | \(\ds c^{-1} g\) | Definition of Identity Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds c^{-1}\) | \(\in\) | \(\ds \map Z G\) | Definition of Center of Group |
$\Box$
Therefore:
- $\map Z G \le G$
$\blacksquare$
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 $G$.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \delta$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.13$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 37.2$ Some important general examples of subgroups
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Examples of groups $\text{(vii)}$