Finite Subgroups

The subgroup criteria show how to verify a subgroup in general. For finite subsets, the verification becomes even shorter. The focus keyword for this lecture is finite subgroup. In a finite closed subset of a group, repeated powers of an element cannot remain distinct forever. This repetition forces the identity and inverse elements to appear. In this class note, we shall prove the finite subgroup theorem and apply it to examples where closure alone determines the subgroup property.

Let

(G,\circ)

be a group and let

H

be a non-empty finite subset of

G

. Then

H

is a subgroup of

G

if and only if

\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}

Let

(G,\circ)

be a group and let

H

be a non-empty finite subset of

G

. Then

H

is a subgroup of

G

if and only if

\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}

Given that

(G,\circ)

is a group and

H

is a non-empty finite subset of

G

. To prove that

H

is a subgroup of

G

if and only if

a\circ b\in H

for all

a,b\in H

. [1] Let

H

be a subgroup of

G

. Then

H

is closed under

\circ

. Therefore

\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}

[2] Let

\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}

To prove that

H

is a subgroup of

G

. Let

a\in H

. Since

H

is closed under

\circ

, we get

\begin{equation*} a,a^2,a^3,\ldots \in H. \end{equation*}

Since

H

is finite, there exist positive integers

r

and

s

such that

r<s

and

\begin{equation*} a^r=a^s. \end{equation*}

Multiplying by

a^{-r}

G

, we get

\begin{align*} a^r=a^s &\implies e=a^{s-r}. \end{align*}

Since

s-r\in\mathbb{N}

and

a^{s-r}\in H

, we get

e\in H

. Now

\begin{align*} a^{s-r}=e &\implies a^{-1}=a^{s-r-1}. \end{align*}

s-r=1

, then

a^{-1}=e\in H

. If

s-r>1

, then

a^{s-r-1}\in H

by closure. Therefore

a^{-1}\in H

. Thus

a^{-1}\in H

for all

a\in H

. By the subgroup criterion,

H

is a subgroup of

G

. Hence,

H

is a subgroup of

G

if and only if

H

is closed under

\circ

\square

The finite condition cannot be removed. For example,

\mathbb{N}

is closed under addition, but it is not a subgroup of

(\mathbb{Z},+)

because additive inverses are missing. In a finite set, closure under the operation traps all repeated powers inside the same finite set, so repetition is unavoidable. That repetition produces the identity and then the inverse. This is the main mathematical reason behind the finite subgroup criterion.

Let

H=\{0,3\}

be a subset of

\mathbb{Z}_6

under addition modulo

6

. Since

H

is finite and non-empty, it is enough to check closure. We compute

\begin{equation*} 0+0\equiv 0,\quad 0+3\equiv 3,\quad 3+0\equiv 3,\quad 3+3\equiv 0\pmod 6. \end{equation*}

Every sum belongs to

H

. Hence,

H

is a subgroup of

\mathbb{Z}_6

Let

K=\{0,2\}

be a subset of

\mathbb{Z}_6

under addition modulo

6

. Then

K

is finite and non-empty. But

\begin{equation*} 2+2\equiv 4\pmod 6 \end{equation*}

and

4\notin K

. Therefore

K

is not closed under addition modulo

6

. Hence,

K

is not a subgroup of

\mathbb{Z}_6

We observe the finite subgroup criterion in

\mathbb{Z}_n

. Choose a finite subset, and the preview checks every possible sum of two elements. In a finite group, if this closure table stays inside the subset, then identity and inverses are forced automatically. This is why the finite theorem needs only closure, while the general subgroup criterion also checks inverses.

The word finite does essential work. For infinite subsets such as

\mathbb{N}\subseteq\mathbb{Z}

, closure under addition does not force inverses.

Let

G

be a group and let

H=\{e,a,a^2\}

, where

a^3=e

. Prove that

H

is a subgroup of

G

Let

(G,\circ)

be a group and let

a\in G

such that

a^3=e

. Let

\begin{equation*} H=\{e,a,a^2\}. \end{equation*}

To prove that

H

is a subgroup of

G

. Since

e\in H

, the set

H

is non-empty. Also

H

is finite. It remains to check closure. The products are governed by the rule

\begin{equation*} a^i\circ a^j=a^{i+j}. \end{equation*}

Since

a^3=e

, every power of

a

reduces to one of

e,a,a^2

. Therefore

\begin{equation*} a^i\circ a^j\in H\quad \forall\,a^i,a^j\in H. \end{equation*}

Thus

H

is closed under

\circ

. By the finite subgroup criterion,

H

is a subgroup of

G

. Hence,

H\leq G

\square

Let

(G,\circ)

be a group such that

\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}

Prove that

H=\{x^2:x\in G\}

is a subgroup of

G

Let

(G,\circ)

be a group such that

\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}

Let

\begin{equation*} H=\{x^2:x\in G\}. \end{equation*}

To prove that

H

is a subgroup of

G

. Since

e=e^2

, we get

e\in H

. Therefore

H

is non-empty. Let

x^2,y^2\in H

, where

x,y\in G

. Using the given condition with

a=y

and

b=x

, we get

\begin{align*} (y\circ x)^3=y^3\circ x^3 &\implies y\circ x\circ y\circ x\circ y\circ x=y\circ y\circ y\circ x\circ x\circ x\\ &\implies x\circ y\circ x\circ y=y^2\circ x^2 \quad [\because \text{ cancellation laws}]. \end{align*}

Interchanging

x

and

y

, we get

\begin{equation*} y\circ x\circ y\circ x=x^2\circ y^2. \end{equation*}

Therefore

\begin{align*} x^2\circ y^2 &=y\circ x\circ y\circ x\\ &=(y\circ x)^2\in H. \end{align*}

Thus

H

is closed under

\circ

. Let

x^2\in H

. Then

\begin{align*} (x^2)^{-1} &=x^{-2}\\ &=(x^{-1})^2\in H. \end{align*}

Therefore

(x^2)^{-1}\in H

for all

x^2\in H

. By the subgroup criterion,

H

is a subgroup of

G

. Hence,

H\leq G

\square

This calculator tests the finite subgroup criterion for subsets of

\mathbb{Z}_n

. Enter a modulus and a subset. The output says whether closure under addition modulo

n

holds and gives the subgroup conclusion when it does. It is a closure calculator, not a general infinite-subset test.

1. Prove that

\{0,2,4\}

is a subgroup of

\mathbb{Z}_6

under addition modulo

6

. 2. Show that

\{0,1,3\}

is not a subgroup of

\mathbb{Z}_4

under addition modulo

4

. 3. Let

a\in G

and

a^4=e

. Prove that

\{e,a,a^2,a^3\}

is a subgroup of

G

1. The set is finite and non-empty. All sums modulo

6

remain in

\{0,2,4\}

, so it is a subgroup. 2. Since

1+1\equiv 2\pmod 4

and

2\notin\{0,1,3\}

, closure fails. 3. Products of elements have the form

a^i a^j=a^{i+j}

, and since

a^4=e

, every product reduces to one of

e,a,a^2,a^3

Because repeated powers must repeat in a finite set, forcing the identity and inverses to appear.

No. The set

\mathbb{N}

is closed under addition but is not a subgroup of

\mathbb{Z}

Yes. A group must contain an identity element.

BMLabs Mathematics

BMLabs Mathematics

Finite Subgroups

Finite Subgroups

FINITE SUBSETS AND CLOSURE

FINITE SUBGROUP CRITERION

PROOF OF FINITE SUBGROUP CRITERION

WHY FINITENESS MATTERS

SUBGROUP OF INTEGERS MODULO SIX

FAILURE OF CLOSURE IN INTEGERS MODULO SIX

EXPLORE CLOSURE IN A FINITE SUBSET

FINITE CLOSURE TABLE EXPLORER

CLOSURE IS ENOUGH ONLY HERE

FINITE CLOSED SUBSET

SOLUTION OF FINITE CLOSED SUBSET

SQUARES UNDER A CUBIC CONDITION

SOLUTION OF SQUARES UNDER A CUBIC CONDITION

CHECK CLOSURE ONLY

FINITE SUBGROUP CLOSURE CALCULATOR

EXERCISES ON FINITE SUBGROUPS

ANSWERS ON FINITE SUBGROUPS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO INTERSECTIONS OF SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation