Quotients by Central Subgroups

After using commutators to recognise abelian quotients, we now focus on quotients by central subgroups. The centre

Z(G)

is always normal, so

G/Z(G)

is always a quotient group. Surprisingly, if this quotient is cyclic, then the original group must be abelian. More generally, if

H

lies inside the centre and

G/H

is cyclic, then

G

is abelian. These results are useful because they turn information about a quotient into information about the original group.

Let

(G,\circ)

be a group. If

G/Z(G)

is cyclic, then

G

is abelian.

After using commutators to recognise abelian quotients, we now focus on quotients by central subgroups. The centre

Z(G)

is always normal, so

G/Z(G)

is always a quotient group. Surprisingly, if this quotient is cyclic, then the original group must be abelian. More generally, if

H

lies inside the centre and

G/H

is cyclic, then

G

is abelian. These results are useful because they turn information about a quotient into information about the original group.

Let

(G,\circ)

be a group. If

G/Z(G)

is cyclic, then

G

is abelian.

Given that

(G,\circ)

is a group and

G/Z(G)

is cyclic. To prove that

G

is abelian. Since

G/Z(G)

is cyclic, there exists

a\in G

such that

\begin{equation*} G/Z(G)=\langle aZ(G)\rangle. \end{equation*}

Let

x,y\in G

. Then there exist integers

m,n

and elements

z_1,z_2\in Z(G)

such that

\begin{equation*} x=a^m\circ z_1 \end{equation*}

and

\begin{equation*} y=a^n\circ z_2. \end{equation*}

Since

z_1,z_2\in Z(G)

, they commute with every element of

G

. Therefore

\begin{align*} x\circ y &=(a^m\circ z_1)\circ(a^n\circ z_2)\\ &=a^m\circ a^n\circ z_1\circ z_2\\ &=a^{m+n}\circ z_1\circ z_2\\ &=a^{n+m}\circ z_2\circ z_1\\ &=a^n\circ a^m\circ z_2\circ z_1\\ &=(a^n\circ z_2)\circ(a^m\circ z_1)\\ &=y\circ x. \end{align*}

Thus

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

Hence,

G

is abelian.

\square

Let

(G,\circ)

be a non-abelian group. Then

G/Z(G)

is not cyclic.

Given that

(G,\circ)

is a non-abelian group. To prove that

G/Z(G)

is not cyclic. If possible let

G/Z(G)

be cyclic. Then, by the previous theorem,

G

is abelian. This contradicts the given fact that

G

is non-abelian. A contradiction. Hence,

G/Z(G)

is not cyclic.

\square

Let

(G,\circ)

be a group and let

H

be a subgroup of

Z(G)

. If

G/H

is cyclic, then

G

is abelian.

Given that

(G,\circ)

is a group,

H

is a subgroup of

Z(G)

, and

G/H

is cyclic. To prove that

G

is abelian. Since

H\subseteq Z(G)

H

is normal in

G

. Since

G/H

is cyclic, there exists

a\in G

such that

\begin{equation*} G/H=\langle aH\rangle. \end{equation*}

Let

x,y\in G

. Then there exist integers

m,n

and elements

h_1,h_2\in H

such that

\begin{equation*} x=a^m\circ h_1 \end{equation*}

and

\begin{equation*} y=a^n\circ h_2. \end{equation*}

Since

H\subseteq Z(G)

, we have

h_1,h_2\in Z(G)

. Therefore

\begin{align*} x\circ y &=(a^m\circ h_1)\circ(a^n\circ h_2)\\ &=a^m\circ a^n\circ h_1\circ h_2\\ &=a^{m+n}\circ h_1\circ h_2\\ &=a^{n+m}\circ h_2\circ h_1\\ &=a^n\circ a^m\circ h_2\circ h_1\\ &=(a^n\circ h_2)\circ(a^m\circ h_1)\\ &=y\circ x. \end{align*}

Thus

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

Hence,

G

is abelian.

\square

Let

G

be a group such that

G/Z(G)

has prime order. Since every group of prime order is cyclic,

G/Z(G)

is cyclic. By the theorem,

G

is abelian. Therefore a non-abelian group cannot have

G/Z(G)

of prime order. Hence, the central quotient of a non-abelian group cannot be cyclic of prime order.

\square

Let

G=Q_8

, the quaternion group. The group

Q_8

is non-abelian, and therefore

\begin{equation*} Q_8/Z(Q_8) \end{equation*}

is not cyclic. In fact, this quotient has four elements and is isomorphic in structure to the Klein four group. This example shows why the theorem is restrictive: a non-abelian group cannot have a cyclic quotient by its centre.

\square

We observe the implication from a cyclic central quotient back to the original group. Choose whether the subgroup being factored out is central and whether the quotient is cyclic. When both facts hold, every element of

G

can be written as a power of one representative times a central element, forcing commutativity. If

G

is known to be non-abelian, the same implication rules out a cyclic quotient by the centre.

Centrality lets the lifted representatives commute after rearranging the central factors. That is why cyclicity of the quotient becomes strong enough to force the whole group to be abelian.

Let

G

be a group such that

|G/Z(G)|=p

, where

p

is prime. Prove that

G

is abelian.

Let

|G/Z(G)|=p

, where

p

is prime. A group of prime order is cyclic. Therefore

G/Z(G)

is cyclic. By the theorem that cyclicity of

G/Z(G)

forces

G

to be abelian, we get

\begin{equation*} G \text{ is abelian}. \end{equation*}

Let

H\leq Z(G)

and suppose

G/H=\langle aH\rangle

. Show directly that every element of

G

has the form

a^n\circ h

for some

n\in\mathbb{Z}

and

h\in H

Let

x\in G

. Since

G/H=\langle aH\rangle

, there exists

n\in\mathbb{Z}

such that

\begin{equation*} xH=(aH)^n. \end{equation*}

Therefore

\begin{equation*} xH=a^nH. \end{equation*}

Thus

\begin{equation*} (a^n)^{-1}\circ x\in H. \end{equation*}

Let

\begin{equation*} h=(a^n)^{-1}\circ x. \end{equation*}

Then

h\in H

, and

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

We combine quotient order and cyclicity facts. Enter the order of

G/Z(G)

if it is finite. If the order is prime, then the quotient is cyclic, and the theorem forces

G

to be abelian. The calculator also warns when a declared non-abelian group is paired with a cyclic central quotient claim.

1. Prove that if

G/Z(G)

is cyclic, then

G

is abelian. 2. Prove that if

G

is non-abelian, then

G/Z(G)

is not cyclic. 3. Let

H\leq Z(G)

. Prove that if

G/H

is cyclic, then

G

is abelian. 4. If

|G/Z(G)|=p

for a prime

p

, prove that

G

is abelian. 5. Explain why

Q_8/Z(Q_8)

cannot be cyclic.

1. If

G/Z(G)=\langle aZ(G)\rangle

, then every

x\in G

has the form

a^m z

with

z\in Z(G)

, and such elements commute. 2. This is the contrapositive of the theorem that cyclic

G/Z(G)

implies abelian

G

. 3. If

G/H=\langle aH\rangle

and

H\leq Z(G)

, then every element has the form

a^n h

with

h\in H

, and such elements commute. 4. Since every group of prime order is cyclic,

G/Z(G)

is cyclic, so

G

is abelian. 5. Since

Q_8

is non-abelian, its quotient by the centre cannot be cyclic.

The centre

Z(G)

is always a normal subgroup of

G

It means more strongly that

G

is abelian, but it does not necessarily say that

G

itself is cyclic.

Elements of a central subgroup commute with everything, so representatives lifted from a cyclic quotient can be rearranged safely.

Yes, but not when the quotient is by the centre in the form

G/Z(G)

BMLabs Mathematics

BMLabs Mathematics

Quotients by Central Subgroups

Quotients by Central Subgroups

OVERVIEW

EXPLORE CENTRAL QUOTIENT CONSEQUENCES

CENTRAL QUOTIENT CONSEQUENCE EXPLORER

WHAT CENTRALITY ADDS

TEST A CENTRAL QUOTIENT CLAIM

CENTRAL QUOTIENT TEST CALCULATOR

FREQUENTLY ASKED QUESTIONS

REVIEW QUOTIENT GROUP CONSTRUCTION

Coordinates and Locus

Invariants under Orthogonal Transformation