Centre of a Group

After studying subgroups formed from intersections, unions, and products, the next natural step is to study subgroups determined by commutativity conditions. The first such special subgroup is the centre of a group. The focus keyword for this lecture is centre of a group. The centre collects exactly those elements that commute with every element of the group. In this lecture, students will learn the definition of the centre, prove that it is a subgroup, and understand why an abelian group is precisely a group whose centre is the whole group.

Let

(G,\circ)

be a group. The \textbf{centre of the group}

G

is the set of all elements of

G

which commute with every element of

G

, that is,

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

Let

(G,\circ)

be a group. The \textbf{centre of the group}

G

is the set of all elements of

G

which commute with every element of

G

, that is,

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

An element

a\in G

belongs to

Z(G)

only when it commutes with every element of

G

, not merely with some elements of

G

. Thus the centre measures how much commutativity is present inside a possibly non-abelian group. If

G

is abelian, then every element commutes with every other element, so the centre should be all of

G

. If

G

is highly non-abelian, the centre may be very small. A common mistake is to check

a\circ x=x\circ a

for one convenient element

x

instead of checking it for all

x\in G

Let

(G,\circ)

be a group. Then

Z(G)

is a subgroup of

G

Given that

(G,\circ)

is a group. To prove that

Z(G)

is a subgroup of

G

. Since

e\circ x=x\circ e=x

for all

x\in G

, we get

e\in Z(G)

. Therefore

Z(G)

is non-empty. Let

a,b\in Z(G)

. To prove that

a\circ b^{-1}\in Z(G)

. Let

x\in G

. Since

a\in Z(G)

, we get

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

Since

b\in Z(G)

, we get

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

Therefore

\begin{align*} b\circ x=x\circ b &\implies b\circ x\circ b^{-1}=x\\ &\implies x\circ b^{-1}=b^{-1}\circ x. \end{align*}

Thus

\begin{equation*} b^{-1}\circ x=x\circ b^{-1}. \end{equation*}

Now

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

Therefore

a\circ b^{-1}\in Z(G)

. By the one-step subgroup criterion,

Z(G)

is a subgroup of

G

. Hence,

Z(G)\leq G

\square

Let

(G,\circ)

be a group. If

G

is abelian, then

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

Given that

(G,\circ)

is a group. Let

G

be abelian. To prove that

Z(G)=G

. Let

a\in G

. Since

G

is abelian, we get

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

Therefore

a\in Z(G)

. Thus

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

Also, by definition of

Z(G)

, we have

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

Therefore

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

Hence,

Z(G)=G

\square

Let

(G,\circ)

be a group. Then

G

is abelian if and only if

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

Given that

(G,\circ)

is a group. To prove that

G

is abelian if and only if

Z(G)=G

. [1] Let

G

be abelian. By the theorem on the centre of an abelian group, we get

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

[2] Let

Z(G)=G

. To prove that

G

is abelian. Let

a,b\in G

. Since

Z(G)=G

, we get

a\in Z(G)

. Therefore

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

Thus every pair of elements of

G

commutes. Hence,

G

is abelian if and only if

Z(G)=G

\square

Let

(\mathbb{Z},+)

be the group of integers under addition. Since addition of integers is commutative, for all

a,x\in\mathbb{Z}

we have

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

Therefore every integer belongs to the centre. Hence,

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

Let

(G,\circ)

be a non-abelian group. Then at least one pair of elements

a,b\in G

satisfies

\begin{equation*} a\circ b\neq b\circ a. \end{equation*}

In that case,

a

cannot belong to

Z(G)

b

cannot belong to

Z(G)

, and hence the centre is not necessarily the whole group. This shows why the centre is a special subgroup measuring the central part of a group rather than the whole group in every case.

We observe centrality in the concrete group

S_3

. Choose an element and test whether it commutes with every element of

S_3

. The identity passes every test, while a transposition or a three-cycle fails with at least one element. This shows why checking one convenient product is not enough to prove membership in the centre.

The centre collects elements that pass every commutativity test. In

S_3

, only the identity passes, which is why the centre of

S_3

is trivial.

Let

G=\{e,a\}

be a group of order

2

. Prove that

Z(G)=G

Let

G=\{e,a\}

be a group of order

2

. To prove that

Z(G)=G

. Since

e

is the identity element, we get

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

Therefore

e\in Z(G)

. Since

a\neq e

and

G

has exactly two elements, the inverse of

a

a

. Thus

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

Now

a

commutes with

e

because

e

is the identity element, and

a

commutes with itself because

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

Therefore

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

Thus

a\in Z(G)

. Hence,

Z(G)=G

\square

We can use the size of the centre to test whether a finite group could be abelian. Enter the order of

G

and the order of

Z(G)

. If the two orders are equal, the abelian centre criterion proves that

G

is abelian. If the centre is smaller, the group is not abelian because an abelian group has centre equal to the whole group.

1. Prove that

e\in Z(G)

for every group

(G,\circ)

. 2. Let

G

be abelian. Prove directly from the definition that

Z(G)=G

. 3. Let

G

be a group. Prove that if

Z(G)=\{e\}

, then no non-identity element of

G

commutes with every element of

G

1. Since

e\circ x=x\circ e=x

for all

x\in G

, we get

e\in Z(G)

. 2. If

a\in G

, then

a\circ x=x\circ a

for all

x\in G

because

G

is abelian. Hence

a\in Z(G)

, so

G\subseteq Z(G)

. Also

Z(G)\subseteq G

, and therefore

Z(G)=G

. 3. If a non-identity element

a

commuted with every element of

G

, then

a\in Z(G)

. This would contradict

Z(G)=\{e\}

Yes. The centre is closed under the one-step subgroup test and contains the identity.

No. The centre is the whole group exactly when the group is abelian.

Yes. Some non-abelian groups have centre equal to

\{e\}

BMLabs Mathematics

BMLabs Mathematics

Centre of a Group

Centre of a Group

CENTRE OF A GROUP

CENTRE OF A GROUP

MEANING OF THE CENTRE

CENTRE IS A SUBGROUP

PROOF OF CENTRE IS A SUBGROUP

CENTRE OF AN ABELIAN GROUP

PROOF OF CENTRE OF AN ABELIAN GROUP

ABELIAN CENTRE CRITERION

PROOF OF ABELIAN CENTRE CRITERION

CENTRE OF AN ADDITIVE GROUP

CENTRE MAY BE PROPER

EXPLORE CENTRAL ELEMENTS

CENTRE MEMBERSHIP EXPLORER

WHAT THE CENTRE MEASURES

CENTRE OF A TWO ELEMENT GROUP

SOLUTION OF CENTRE OF A TWO ELEMENT GROUP

CHECK CENTRE SIZE AND ABELIANITY

CENTRE SIZE CALCULATOR

EXERCISES ON CENTRE OF A GROUP

ANSWERS ON CENTRE OF A GROUP

FREQUENTLY ASKED QUESTIONS

CONTINUE TO CENTRALISERS

Coordinates and Locus

Invariants under Orthogonal Transformation