The Center of a Group

In a general group, elements do not necessarily commute. We can have $ab ≠ ba$.

The elements that commute with every element of the group are special. They form what is called the center of the group.

For a group $G$,

$Z(G) = \{z ∈ G : zx = xz \ \text{ for all } x ∈ G\}$

So an element is in the center if it commutes with every element of $G$.

The identity is always in the center because $ex = xe = x$ for every $x ∈ G$. Therefore the center is never empty.

What Happens in an Abelian Group?

If $G$ is abelian, every pair of elements commutes. That means every element belongs to the center, so $Z(G) = G$.

The converse is also true. If every element of $G$ lies in the center, then every pair of elements commutes, so $G$ is abelian.

Therefore: $G \text{ is abelian } ⇔ Z(G) = G$

Example: $S_3$

The symmetric group $S_3$ has six elements and is nonabelian.

For example, $(12)(23) = (123)$ but $(23)(12) = (132)$.

So those elements do not commute. In fact, no nonidentity element of $S_3$ commutes with every element of $S_3$. Therefore $Z(S_3) = \{e\}$.

A group whose center contains only the identity is said to have a trivial center.

Example: $D_4$

Now consider $D_4$, the symmetry group of a square.

A rotation does not commute with every reflection, so $D_4$ is not abelian. But the $180°$ rotation does commute with every symmetry of the square.

Therefore $Z(D_4) = \{e, r²\}.$ So a nonabelian group can still have a nontrivial center.

Example: $Q_8$

For the quaternion group $Q_8 = \{±1, ±i, ±j, ±k\},$the elements $1$ and $−1$ commute with every element. The elements $±i, ±j$, and $±k$ do not.

Thus $Z(Q_8) = \{1, −1\}$.

The Center Is Always a Subgroup

The center is not just a set of special elements. It is always a subgroup of $G$: $Z(G) ≤ G$.

Using the subgroup test, suppose $a$ and $b$ are in $Z(G)$. Since $b$ is in the center, $b⁻¹$ also commutes with every $x ∈ G$. Then

$(ab⁻¹)x = a(b⁻¹x) = a(xb⁻¹) = (ax)b⁻¹ = (xa)b⁻¹ = x(ab⁻¹).$

Therefore $ab⁻¹$ is also in $Z(G)$, so $Z(G)$ is a subgroup.

The Center Is Normal

In fact, the center is always a normal subgroup: $Z(G) ◁ G.$

If $z ∈ Z(G)$and $g ∈ G$, then $z$ commutes with $g$: $gz = zg$.

Multiplying on the right by $g⁻¹$ gives $gzg⁻¹ = z$. So conjugating an element of the center leaves it in the center. Therefore $Z(G)$ is normal in $G$.

Watch the Video

If you want to see the related idea for a single element, read What Is the Centralizer of an Element? https://dogmathic.com/centralizer/

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