Center is Normal Subgroup
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Theorem
The center $Z \left({G}\right)$ of any group $G$ is a normal subgroup of $G$ which is abelian.
Proof 1
We have the result Center is Abelian Subgroup.
Since $g x = x g$ for each $g \in G$ and $x \in Z \left({G}\right)$, we have $g Z \left({G}\right) = Z \left({G}\right) g$.
Thus, $Z \left({G}\right) \triangleleft G$.
$\blacksquare$
Proof 2
We have:
- $\forall a \in G: x \in Z \left({G}\right)^a \iff a x a^{-1} = x a a^{-1} = x \in Z \left({G}\right)$
Therefore:
- $\forall a \in G: Z \left({G}\right)^a = Z \left({G}\right)$
and $Z \left({G}\right)$ is a normal subgroup of $G$.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $12.11 \ \text{(a)}$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 46 \theta$
- John F. Humphreys: A Course in Group Theory (1996)... (previous)... (next): $\S 7$: Exercise $5$