The Group Theory Society

01

THE FOUR OATHS

No oath, no group.

When proving a newly defined operation forms a group, all four requirements must be addressed. The society does not accept ‘it feels group-ish.’

01↳

Closure

\forall x,y\in G,\quad x*y\in G

What happens in G stays in G.

Open field notes ↘

PLAIN LANGUAGECombine two members and the result cannot escape the set.

SHORT EXAMPLEIntegers are closed under addition: 3 + (−8) = −5 ∈ ℤ.

EXAM WORDINGLet x,y ∈ G. Since …, x ∗ y ∈ G. Thus ∗ is closed on G.

FREQUENT MISTAKEProving the result exists but not proving it lies in G.

02( )

Associativity

\forall x,y,z\in G,\quad (x*y)*z=x*(y*z)

Move the parentheses, not the elements.

Open field notes ↘

PLAIN LANGUAGEYou may move the grouping marks while keeping the actors in exact order.

SHORT EXAMPLEFor addition, (2 + 3) + 4 = 2 + (3 + 4) = 9.

EXAM WORDINGLet x,y,z ∈ G. Compute both bracketings and simplify them to the same expression.

FREQUENT MISTAKESwapping x and y. That is commutativity, a completely different oath.

03e

Identity

\exists e\in G:\quad x*e=e*x=x

Does nothing. Still essential. Iconic.

Open field notes ↘

PLAIN LANGUAGEOne element changes absolutely nothing—on both sides.

SHORT EXAMPLE0 is the identity in (ℤ,+); 1 is the identity in (ℝ*,×).

EXAM WORDINGSolving x ∗ e = x gives e = …. Verify also e ∗ x = x.

FREQUENT MISTAKEChecking only the right identity in an operation not known to commute.

04↶

Inverses

\forall x\in G,\ \exists x^{-1}\in G:\quad x*x^{-1}=x^{-1}*x=e

Every element’s mathematical undo button.

Open field notes ↘

PLAIN LANGUAGEEvery member has a legal undo move that returns it to identity.

SHORT EXAMPLEThe additive inverse of 7 is −7 because 7 + (−7) = 0.

EXAM WORDINGFor arbitrary x, solve x ∗ y = e, show y ∈ G, then verify y ∗ x = e.

FREQUENT MISTAKEFinding a candidate but never checking it belongs to G or works both ways.

EXAM TEMPLATE · COPY THIS RHYTHM

\forall x,y\in G,\quad x*y\in G

\forall x,y,z\in G,\quad (x*y)*z=x*(y*z)

\exists e\in G:\quad x*e=e*x=x

\forall x\in G,\ \exists x^{-1}\in G:\quad x*x^{-1}=x^{-1}*x=e

(G,*)\text{ is a group.}

02

PARENTHESES VS. ORDER

Regrouping is legal. Reordering is a felony.

Associativity and commutativity answer entirely different questions. A noncommutative group holds grudges about order.

ASSOCIATIVITY

Move the brackets.

(x*y)*z=x*(y*z)

Changes grouping. The sequence x, y, z is sacred.

≠

COMMUTATIVITY

Swap the elements.

x*y=y*x

Changes order. One counterexample disproves it.

INTERACTIVE · DRAG BY BUTTON

Parenthesis Transport Authority

Parentheses have travel documents. Elements do not.

(x∗y)∗(z)

Enable commutativityOrder locked. This group remembers.

03

NOTHING & UNDO

Iconic neutrality. Elegant cancellation.

Find candidates algebraically, then verify them on both sides. Unknown operations have not earned your trust yet.

e

IDENTITY PROTOCOL

Find the element that leaves x untouched.

x*e=x

Solve for the unknown e, then verify the other direction:

e*x=x

x∗ ex

xx⁻¹e

INVERSE PROTOCOL

Cancel yourself back home.

x*y=e

Solve for y, prove y belongs to G, then reverse the order:

y*x=e

04

THE CHILD SUBGROUP

Membership has standards.

H must be a subset of G using the same operation. Landing somewhere in G is not enough—the result must return to H.

FULL TEST · THREE CHECKS

01 H is closed.
02 $e\in H$ .
03 Every inverse remains in H.

FAST TEST · ONE MOVE

H\ne\varnothing

\forall a,b\in H,\quad ab^{-1}\in H

For additive groups: $a-b\in H$ .

CRITICAL LANDING ZONE~~ab⁻¹ ∈ G~~ → ab⁻¹ ∈ H

The result must land inside H, not merely inside G.

MINI GAME

Subgroup Bouncer

Which candidate gets through all three doors?

H MEMBERS ONLY

Choose an applicant.

“First, $H\ne\varnothing$ . Let $a,b\in H$ . We have $ab^{-1}=\cdots\in H$ . Hence, by the subgroup criterion, $H\leq G$ .”

05

THE NORMALITY RITUAL

The mother wraps the child—and sends it home.

Normality is extra status. First prove H is a subgroup; then test every outer g from G and inner h from H.

ghg⁻¹→ghg⁻¹ ∈ H

Outer g comes from mother group G

Inner h comes from child subgroup H

Result must return to H

CONJUGATION TEST

\forall g\in G,\ \forall h\in H,\quad ghg^{-1}\in H

G — H — G⁻¹

EQUIVALENT COSET TEST

gH=Hg\quad\forall g\in G

Every left coset agrees with its right coset.

WHY ABELIAN SUBGROUPS ARE AUTOMATICALLY NORMAL

ghg^{-1}=hgg^{-1}=h

The rearrangement is permitted because the group is abelian. In a nonabelian group, that swap is illegal.

“Since $H\leq G$ , let $g\in G$ and $h\in H$ . Then $ghg^{-1}=\cdots\in H$ . Therefore $H\trianglelefteq G$ .”

06

THE UNIVERSE MUST AGREE

Two paths. One output.

A homomorphism preserves the operation for arbitrary elements. One lucky numerical example is evidence, not proof.

f:(G,*)\to(K,\circ)

\forall x,y\in G,\quad f(x*y)=f(x)\circ f(y)

“Operation first, then map = map first, then operation.”

(a,b) is one group elementTwo coordinates do not make two elements. Coordinate identity crisis averted.

(x,y)

x ∗ y

f(x), f(y)

same output

operate → mapmap → operate

PROJECTION FIELD NOTE

$f(a,b)=b$

If $x=(a,b)$ and $y=(c,d)$ , then $f(x)=b$ and $f(y)=d$ . The ordered pairs are the inputs; b and d are their images.

“Let $x,y\in G$ . We compute $f(x*y)=\cdots$ , while $f(x)\circ f(y)=\cdots$ . Therefore $f(x*y)=f(x)\circ f(y)$ , so f is a group homomorphism.”

07

AUTHORITATIVE EXAM TRANSCRIPTION

Exercise 2 · 45 points

Photographed Lebanese University final exam, First Semester, M 2250. Every Group Theory question below preserves the original numbering, definitions, symbols, and wording.

SOURCE LOCKED · 2024–2025 FIRST SEMESTER

Exercise 2 (45 points) Let ∗ be the binary operation defined on $G=\mathbb R\times\mathbb R^*$ by:

(a,b)*(c,d)=(a+bc,bd).

1. Show that:

∗ is associative.
∗ is not commutative.
∗ has a neutral element to be determined.
Calculate $(a,b)*\left(-\frac ab,\frac1b\right)$
Deduce that $(G,*)$ is a group.

2. Let $H=\{(a,b)\in G/b>0\}$

Show that H is a subgroup of $(G,*)$ .
Is H a normal subgroup of $(G,*)$ ? Justify your answer.

3. Is the set $K=\{(a,b)\in G/a<0\text{ and }b<0\}$ a subgroup of $(G,*)$ ? Justify your answer.

4. Let

f:(G,*)\to(\mathbb R^*,\times),\qquad (a,b)\mapsto f(a,b)=b

Show that f is a group homomorphism.

OPERATION CALCULATOR

Order matters. This group holds grudges.

abcd

$(a,b)*(c,d)$ (2, 2)

$(c,d)*(a,b)$ (1, 2)

NOT EQUAL · noncommutativity witnessed

VISUAL ASSOCIATIVITY CHECKER

$((a,b)*(c,d))*(e,f)$

(a + bc + bde, bdf)

=

$(a,b)*((c,d)*(e,f))$

(a + bc + bde, bdf)

PROOF PRECISION CHECKPOINTS

✓ Closure must be explicitly established.

✓ The identity should be verified on both sides.

✓ The inverse should be verified on both sides.

✓ In a noncommutative group, factors cannot be reordered.

✓ To prove membership in H, the entire ordered pair must belong to H.

✓ Arbitrary elements—not one example—must be used in universal proofs.

1(a)

∗ is associative.

1(b)

∗ is not commutative.

1(c)

∗ has a neutral element to be determined.

1(d)

Calculate $(a,b)*\left(-\frac{a}{b},\frac1b\right)$ .

1(e)

Deduce that $(G,*)$ is a group.

2(a)

Show that H is a subgroup of $(G,*)$ .

2(b)

Is H a normal subgroup of $(G,*)$ ? Justify your answer.

3

Is the set $K=\{(a,b)\in G/a<0\text{ and }b<0\}$ a subgroup of $(G,*)$ ? Justify your answer.

4

Let $f:(G,*)\to(\mathbb R^*,\times),\quad (a,b)\mapsto f(a,b)=b$ . Show that f is a group homomorphism.

08

HALL OF ACADEMIC DISASTERS

The Mistake Museum

Observe the wild proof error in its natural habitat. Learn why it fails. Do not tap the glass.

EXHIBIT 01

The Wrong Landing

✕ ab⁻¹ ∈ G

The subgroup test needs ab⁻¹ ∈ H. G is the lobby; H is the destination.

CORRECTED

EXHIBIT 02

The Inner-Circle Coup

✕ g,h ∈ H in normality

Use g ∈ G and h ∈ H. The outer element represents the whole mother group.

CORRECTED

EXHIBIT 03

Premature Normality

✕ Proving normality first

Normality is additional status. First prove H ≤ G.

CORRECTED

EXHIBIT 04

The Illegal Swap

✕ ghg⁻¹ = hgg⁻¹ always

Only abelian groups issue that reordering permit.

CORRECTED

EXHIBIT 05

One Example Fallacy

✕ One pair proves f is a homomorphism

A universal statement requires arbitrary x,y ∈ G.

CORRECTED

EXHIBIT 06

Coordinate Identity Crisis

✕ (a,b) means two elements

It is one element with two coordinates.

CORRECTED

EXHIBIT 07

The One-Sided Celebrity

✕ Only x∗e=x checked

Verify e∗x=x too unless commutativity is known.

CORRECTED

EXHIBIT 08

The Missing Oath

✕ Associative + identity + inverses

Closure has quietly left the building. Prove it.

CORRECTED

EXHIBIT 09

Scalar Trespassing

✕ b ∈ H

H contains ordered pairs. Say (u,b) ∈ H after checking its defining condition.

CORRECTED

EXHIBIT 10

Parenthesis Fraud

✕ Associative = commutative

Associative regroups; commutative reorders. Different crimes.

CORRECTED

09

FINAL BOSS

Prove your membership.

Fifteen shuffled trials. Instant explanations. One dramatically judgmental rank.

Multiple choice

TRIAL 1 / 15

Which line proves closure?

10

BREAK GLASS BEFORE EXAM

Last-Minute Cheat Sheet

The shortest route through the ritual. Print it, annotate it, keep it away from actual exam desks.

GH

M2250 · RAPID RECALL

GROUP

\text{Closure + Associativity + Identity + Inverses}

All four. No exceptions.

SUBGROUP

H\ne\varnothing,\quad a,b\in H\Rightarrow ab^{-1}\in H

The answer lands in H.

NORMAL SUBGROUP

H\leq G,\quad g\in G,\ h\in H\Rightarrow ghg^{-1}\in H

Mother · child · mother⁻¹.

HOMOMORPHISM

f(x*y)=f(x)\circ f(y)

Operate then map = map then operate.

ASSOCIATIVITY

(x*y)*z=x*(y*z)

Regroup only.

COMMUTATIVITY

x*y=y*x

Reorder elements.

ENTER THEWOMB GROUP.

Closure

Associativity

Identity

Inverses

Move the brackets.

Swap the elements.

Parenthesis Transport Authority

Find the element that leaves x untouched.

Cancel yourself back home.

Subgroup Bouncer

f(a,b)=bf(a,b)=bf(a,b)=b

Order matters. This group holds grudges.

The Wrong Landing

The Inner-Circle Coup

Premature Normality

The Illegal Swap

One Example Fallacy

Coordinate Identity Crisis

The One-Sided Celebrity

The Missing Oath

Scalar Trespassing

Parenthesis Fraud

Which line proves closure?

THE GROUP THEORY SOCIETY

ENTER THE
WOMB GROUP.

$f(a,b)=b$