# Center of Group is Normal Subgroup

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## Theorem

Although this article appears correct, it's inelegant. There has to be a better way of doing it.Title and statement slightly differ, should we remove "which is abelian"?You can help Proof Wiki by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

The center $\map Z G$ of any group $G$ is a normal subgroup of $G$ which is abelian.

## 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: $\S 46 \theta$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Exercise $5$