Center of Group is Normal Subgroup

Theorem

Let $G$ be a group

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


Proof 1

Recall that Center of Group is Abelian Subgroup.

Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$:

$g \map Z G = \map Z G g$

Thus:

$\map Z G \lhd G$

$\blacksquare$


Proof 2

We have:

$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$

Therefore:

$\forall a \in G: \map Z G^a = \map Z G$

and $\map Z G$ is a normal subgroup of $G$.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $10$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.11 \ \text{(a)}$
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \theta$
  • 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): normal subgroup or invariant subgroup
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Exercise $5$
  • 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): centre (group theory)
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