Centre and Centralisers

After defining centralisers, we now connect them with the centre of a group. The focus keyword for this lecture is centre and centralisers. The centre consists of elements that commute with all elements, while a centraliser consists of elements that commute with one fixed element. This difference immediately gives an important inclusion: the centre is contained in every centraliser. In this lecture, students will prove this inclusion, express the centre as an intersection of centralisers, and study the condition

C(a)=Z(G)

Let

(G,\circ)

be a group and let

a\in G

. Then

Z(G)

is a subgroup of

C(a)

C(a)=Z(G)

Let

(G,\circ)

be a group and let

a\in G

. Then

Z(G)

is a subgroup of

C(a)

Given that

(G,\circ)

is a group and

a\in G

. To prove that

Z(G)

is a subgroup of

C(a)

. Since

Z(G)

is a subgroup of

G

, it is enough to prove that

\begin{equation*} Z(G)\subseteq C(a). \end{equation*}

Let

x\in Z(G)

. Then

\begin{equation*} x\circ g=g\circ x\quad \forall\,g\in G. \end{equation*}

Since

a\in G

, we get

\begin{equation*} x\circ a=a\circ x. \end{equation*}

Therefore

x\in C(a)

. Thus

\begin{equation*} Z(G)\subseteq C(a). \end{equation*}

Since

Z(G)\leq G

and

C(a)\leq G

, the subset relation gives

Z(G)\leq C(a)

. Hence,

Z(G)

is a subgroup of

C(a)

\square

Let

(G,\circ)

be a group. Then

\begin{equation*} Z(G)=\bigcap_{a\in G}C(a). \end{equation*}

Given that

(G,\circ)

is a group. To prove that

Z(G)=\bigcap_{a\in G}C(a)

. Let

x\in Z(G)

. Then

\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}

Therefore

\begin{equation*} x\in C(a)\quad \forall\,a\in G. \end{equation*}

Thus

\begin{equation*} x\in \bigcap_{a\in G}C(a). \end{equation*}

Hence

\begin{equation*} Z(G)\subseteq \bigcap_{a\in G}C(a). \end{equation*}

Conversely, let

x\in \bigcap_{a\in G}C(a)

. Then

\begin{equation*} x\in C(a)\quad \forall\,a\in G. \end{equation*}

Therefore

\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}

Thus

x\in Z(G)

. Hence

\begin{equation*} \bigcap_{a\in G}C(a)\subseteq Z(G). \end{equation*}

Therefore

\begin{align*} Z(G)\subseteq \bigcap_{a\in G}C(a)\quad \text{and}\quad \bigcap_{a\in G}C(a)\subseteq Z(G) &\implies Z(G)=\bigcap_{a\in G}C(a). \end{align*}

Hence,

\begin{equation*} Z(G)=\bigcap_{a\in G}C(a). \end{equation*}

\square

Let

(G,\circ)

be a group. If there exists an element

a\in G

such that

\begin{equation*} C(a)=Z(G), \end{equation*}

then

G

is abelian.

Given that

(G,\circ)

is a group. Let there exist an element

a\in G

such that

\begin{equation*} C(a)=Z(G). \end{equation*}

To prove that

G

is abelian. Since

\begin{equation*} a\circ a=a\circ a, \end{equation*}

we get

a\in C(a)

. Using

C(a)=Z(G)

, we get

a\in Z(G)

. Since

a\in Z(G)

, we get

\begin{equation*} a\circ x=x\circ a\quad \forall\,x\in G. \end{equation*}

Therefore every element of

G

commutes with

a

. Thus

\begin{equation*} G\subseteq C(a). \end{equation*}

Also

\begin{equation*} C(a)\subseteq G. \end{equation*}

Therefore

\begin{equation*} C(a)=G. \end{equation*}

Using

C(a)=Z(G)

, we get

\begin{equation*} Z(G)=G. \end{equation*}

By the abelian centre criterion,

G

is abelian. Hence,

G

is abelian.

\square

The condition

C(a)=Z(G)

is stronger than it may first appear. Since

a

always belongs to

C(a)

, equality forces

a

to lie in the centre. But once

a

is central, every element of the group commutes with

a

, so

C(a)=G

. Therefore

C(a)=Z(G)

forces

Z(G)=G

, and this makes the whole group abelian. This argument is a useful example of how a local commutativity condition can become a global commutativity conclusion.

Let

(G,\circ)

be an abelian group and let

a\in G

. Then

\begin{equation*} Z(G)=G \end{equation*}

and

\begin{equation*} C(a)=G. \end{equation*}

Thus

\begin{equation*} C(a)=Z(G)=G. \end{equation*}

This example shows that the equality

C(a)=Z(G)

occurs naturally in abelian groups.

Let

(G,\circ)

be a group and suppose

\begin{equation*} C(a)=G \end{equation*}

for every

a\in G

. Prove that

G

is abelian.

Let

(G,\circ)

be a group and suppose

C(a)=G

for every

a\in G

. To prove that

G

is abelian. Let

x,y\in G

. Since

C(y)=G

, we get

x\in C(y)

. Therefore

\begin{equation*} x\circ y=y\circ x. \end{equation*}

Thus every pair of elements of

G

commutes. Hence,

G

is abelian.

\square

Let

(G,\circ)

be a group. Prove that if

x

belongs to every centraliser

C(a)

with

a\in G

, then

x\in Z(G)

Let

(G,\circ)

be a group. Suppose that

x\in C(a)

for every

a\in G

. To prove that

x\in Z(G)

. Since

x\in C(a)

for every

a\in G

, we get

\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}

By the definition of the centre, this means

x\in Z(G)

. Hence,

x\in Z(G)

\square

We observe the formula

Z(G)=\bigcap_{a\in G}C(a)

by tracking whether one candidate element belongs to several centralisers. A candidate is central only when it belongs to every centraliser. The display is deliberately logical rather than computational: missing even one centraliser means the element fails to commute with at least one group element. This helps separate a local centraliser condition from the global centre condition.

The centre is the common part of all centralisers. Belonging to one centraliser only proves commutativity with one fixed element; belonging to all centralisers proves commutativity with every element.

We use orders to test the special condition

C(a)=Z(G)

. Since

Z(G)\subseteq C(a)

, equal finite orders force equality. The theorem then says that if such an equality occurs for some element

a

, the whole group must be abelian. The calculator reports this conclusion and warns when the entered orders contradict the required containment.

1. Prove that

Z(G)\subseteq C(a)

for every

a\in G

. 2. Prove that if

C(a)=G

for some

a\in G

, then

a\in Z(G)

. 3. Prove that

G

is abelian if and only if

C(a)=G

for every

a\in G

1. If

x\in Z(G)

, then

x

commutes with every element of

G

, so in particular

x\circ a=a\circ x

. Hence

x\in C(a)

. 2. Since

C(a)=G

, every

x\in G

belongs to

C(a)

. Thus

x\circ a=a\circ x

for every

x\in G

, and hence

a\in Z(G)

. 3. If

G

is abelian, then every element commutes with every

a\in G

, so

C(a)=G

. Conversely, if

C(a)=G

for every

a\in G

, then every pair of elements commutes, so

G

is abelian.

A central element commutes with every element, so it commutes with the particular element

a

It means that an element is central exactly when it belongs to the centraliser of every element of the group.

Since

a\in C(a)

, equality gives

a\in Z(G)

. Then

C(a)=G

, so

Z(G)=G

, and therefore

G

is abelian.

BMLabs Mathematics

BMLabs Mathematics

Centre and Centralisers

Centre and Centralisers

CENTRE AND CENTRALISERS

CENTRE LIES IN EVERY CENTRALISER

PROOF OF CENTRE LIES IN EVERY CENTRALISER

CENTRE AS AN INTERSECTION OF CENTRALISERS

PROOF OF CENTRE AS AN INTERSECTION OF CENTRALISERS

CENTRALISER EQUAL TO CENTRE

PROOF OF CENTRALISER EQUAL TO CENTRE

HOW TO READ THE EQUALITY CONDITION

ABELIAN CASE

FINDING THE CENTRE FROM CENTRALISERS

SOLUTION OF FINDING THE CENTRE FROM CENTRALISERS

USING THE INTERSECTION FORMULA

SOLUTION OF USING THE INTERSECTION FORMULA

EXPLORE CENTRE AS AN INTERSECTION

CENTRALISER INTERSECTION EXPLORER

WHY THE INTERSECTION FORMULA MATTERS

TEST EQUALITY OF CENTRE AND CENTRALISER

CENTRE CENTRALISER EQUALITY CALCULATOR

EXERCISES ON CENTRE AND CENTRALISERS

ANSWERS ON CENTRE AND CENTRALISERS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO COSETS OF SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation