Centraliser of an Element

After studying the centre of a group, we now weaken the commutativity requirement. The centre asks an element to commute with every element of the group, while the centraliser asks which elements commute with one fixed element. The focus keyword for this lecture is centraliser of an element. This subgroup is important because it isolates the part of the group that behaves commutatively around a chosen element. In this lecture, students will define the centraliser, prove that it is a subgroup, and compute centralisers in standard situations.

Let

(G,\circ)

be a group and let

a\in G

. The \textbf{centraliser of the element}

a

G

is the set of all elements of

G

which commute with

a

, that is,

\begin{equation*} C(a)=\{x\in G:x\circ a=a\circ x\}. \end{equation*}

Let

(G,\circ)

be a group and let

a\in G

. The \textbf{centraliser of the element}

a

G

is the set of all elements of

G

which commute with

a

, that is,

\begin{equation*} C(a)=\{x\in G:x\circ a=a\circ x\}. \end{equation*}

The centraliser

C(a)

depends on the chosen element

a

. If

a

changes, the subgroup

C(a)

may also change. The centre

Z(G)

is stricter because an element in the centre must commute with every element of

G

. Thus every central element belongs to every centraliser, but an element may belong to one centraliser without being central. A common mistake is to think that

x\in C(a)

means

x

commutes with all elements of

G

; it means only that

x

commutes with the fixed element

a

Let

(G,\circ)

be a group and let

a\in G

. Then

C(a)

is a subgroup of

G

Given that

(G,\circ)

is a group and

a\in G

. To prove that

C(a)

is a subgroup of

G

. Since

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

we get

e\in C(a)

. Therefore

C(a)

is non-empty. Let

x,y\in C(a)

. Then

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

and

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

From

y\circ a=a\circ y

, we get

\begin{align*} y\circ a=a\circ y &\implies y\circ a\circ y^{-1}=a\\ &\implies a\circ y^{-1}=y^{-1}\circ a. \end{align*}

Thus

\begin{equation*} y^{-1}\circ a=a\circ y^{-1}. \end{equation*}

Now

\begin{align*} (x\circ y^{-1})\circ a &=x\circ(y^{-1}\circ a)\\ &=x\circ(a\circ y^{-1}) \quad [\because y^{-1}\circ a=a\circ y^{-1}]\\ &=(x\circ a)\circ y^{-1}\\ &=(a\circ x)\circ y^{-1} \quad [\because x\circ a=a\circ x]\\ &=a\circ(x\circ y^{-1}). \end{align*}

Therefore

x\circ y^{-1}\in C(a)

. By the one-step subgroup criterion,

C(a)

is a subgroup of

G

. Hence,

C(a)\leq G

\square

Let

(G,\circ)

be an abelian group and let

a\in G

. Since every element of

G

commutes with every other element of

G

, we have

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

Therefore every element of

G

belongs to

C(a)

. Hence,

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

Let

(G,\circ)

be a group with identity element

e

. For every

x\in G

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

Therefore every element of

G

commutes with

e

. Hence,

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

Let

(G,\circ)

be a group and let

a\in G

. Prove that

\langle a\rangle\subseteq C(a)

Let

(G,\circ)

be a group and let

a\in G

. To prove that

\langle a\rangle\subseteq C(a)

. Let

x\in\langle a\rangle

. Then there exists

n\in\mathbb{Z}

such that

\begin{equation*} x=a^n. \end{equation*}

Now

\begin{align*} x\circ a &=a^n\circ a\\ &=a^{n+1}\\ &=a\circ a^n\\ &=a\circ x. \end{align*}

Therefore

x\in C(a)

. Thus

\langle a\rangle\subseteq C(a)

. Hence, every power of

a

belongs to the centraliser of

a

\square

Let

(G,\circ)

be a group and let

a\in G

such that

a^2=e

. Prove that

a\in C(a)

and

a^{-1}\in C(a)

Let

(G,\circ)

be a group and let

a\in G

such that

a^2=e

. To prove that

a\in C(a)

and

a^{-1}\in C(a)

. Since

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

we get

a\in C(a)

. Since

a^2=e

, we get

a^{-1}=a

. Therefore

\begin{equation*} a^{-1}=a\in C(a). \end{equation*}

Hence,

a\in C(a)

and

a^{-1}\in C(a)

\square

The centraliser should be read as a local commutativity subgroup. It does not claim that the whole group is abelian. It only records those elements that commute with one fixed element. In many later topics, especially conjugacy and class equations, the size of

C(a)

tells us how much freedom remains after fixing the element

a

. This is why centralisers appear naturally in structural questions about non-abelian groups.

We observe centralisers in

S_3

. Choose the fixed element

a

, and the display lists exactly the elements

x

satisfying

xa=ax

. The identity has the whole group as its centraliser, while a transposition has a smaller centraliser containing only itself and the identity. This makes the difference between the centre and a centraliser visible: a centraliser is local to one element.

The centraliser always contains the fixed element and its powers. It may be much larger, but it does not have to be the whole group unless the fixed element commutes with every element.

In a finite group,

C(a)

is a subgroup, so its order must divide

|G|

. It also contains

\langle a\rangle

, so

|\langle a\rangle|

must divide

|C(a)|

. Enter the three orders to test these necessary conditions. Passing the divisibility checks does not compute the centraliser, but it confirms that the proposed sizes are compatible with the subgroup facts in this lesson.

1. Prove that

a\in C(a)

for every

a\in G

. 2. Prove that if

G

is abelian, then

C(a)=G

for every

a\in G

. 3. Let

x\in C(a)

. Prove that

x^{-1}\in C(a)

1. Since

a\circ a=a\circ a

, we get

a\in C(a)

. 2. If

G

is abelian, then

x\circ a=a\circ x

for every

x\in G

. Hence

C(a)=G

. 3. Since

x\in C(a)

, we have

x\circ a=a\circ x

. This gives

x^{-1}\circ a=a\circ x^{-1}

, and therefore

x^{-1}\in C(a)

Yes. Every element commutes with itself, so

a\in C(a)

Yes. It satisfies the one-step subgroup criterion.

Not always. The centre contains elements that commute with every element, while

C(a)

contains elements that commute with the fixed element

a

BMLabs Mathematics

BMLabs Mathematics

Centraliser of an Element

Centraliser of an Element

CENTRALISER OF AN ELEMENT

CENTRALISER OF AN ELEMENT

CENTRALISER COMPARED WITH CENTRE

CENTRALISER IS A SUBGROUP

PROOF OF CENTRALISER IS A SUBGROUP

CENTRALISER IN AN ABELIAN GROUP

CENTRALISER OF THE IDENTITY

CENTRALISER CONTAINS POWERS

SOLUTION OF CENTRALISER CONTAINS POWERS

CENTRALISER OF AN ELEMENT OF ORDER TWO

SOLUTION OF CENTRALISER OF AN ELEMENT OF ORDER TWO

CENTRALISER AS A LOCAL COMMUTATIVE PART

EXPLORE CENTRALISERS IN A CONCRETE GROUP

CENTRALISER IN S3 EXPLORER

WHAT THE CENTRALISER RECORDS

CHECK CENTRALISER ORDERS

CENTRALISER ORDER CALCULATOR

EXERCISES ON CENTRALISERS

ANSWERS ON CENTRALISERS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO CENTRE AND CENTRALISERS

Coordinates and Locus

Invariants under Orthogonal Transformation