Groups (Further Pure 2, section 1) — 9FM0 notes

1.1

Year 1

The Axioms of a group.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A group is a set $G$ together with a binary operation $*$ satisfying four axioms. Closure: $a*b\in G$ for all $a,b\in G$ .
Associativity: $(a*b)*c=a*(b*c)$ for all $a,b,c\in G$ . Identity: there is an $e\in G$ with $e*a=a*e=a$ for every $a\in G$ .
Inverses: for every $a\in G$ there is an $a^{-1}\in G$ with $a*a^{-1}=a^{-1}*a=e$ . Commutativity is not an axiom; a group in which $a*b=b*a$ always holds is called abelian.
When you are asked to verify the axioms for a finite set, closure is read off the Cayley table, associativity is usually quoted from the underlying arithmetic, and the identity and inverses are found row by row.
When you are asked to show that a set is not a group, one counterexample to one axiom is enough, and you should name which axiom fails.

Worked example

Show that $G=\{1,5,7,11\}$ under multiplication modulo $12$ is a group.

1.Closure: $5\times7=35\equiv11$ , $5\times11=55\equiv7$ , $7\times11=77\equiv5$ , and each element squared gives $1$ , so every product lies in $G$ .
2.Associativity is inherited from multiplication of integers, which is associative, and taking remainders preserves it.
3.Identity: $1\times a\equiv a$ for every $a\in G$ , so $e=1$ .
4.Inverses: $5\times5\equiv1$ , $7\times7\equiv1$ and $11\times11\equiv1$ , so every element is its own inverse.

Answer: All four axioms hold, so $G$ is a group of order $4$ (and it is abelian).

Common mistakes

Don't fall into the trap of claiming associativity holds because a few products were checked; quote the underlying operation instead.
Don't fall into the trap of forgetting to state that the identity must satisfy $e*a=a*e=a$ on both sides.
Don't fall into the trap of saying a set is a group without checking that inverses stay inside the set.

Exam tip

Name the axiom you are testing before each line of working; marks are awarded axiom by axiom, and a single named counterexample earns the disproof marks.

Tier 1 · Easy

ORIGINAL

1.

State the four axioms that a set

G

with a binary operation

*

must satisfy to be a group.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Determine, with justification, whether the set of all

2\times2

matrices with real entries forms a group under matrix multiplication.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that the identity element of a group is unique.

(4)

(Total for Question 1 is 4 marks)

1.2

Year 1

Examples of groups. Cayley tables. Cyclic groups.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

The standard examples are the integers modulo $n$ under addition, the non-zero residues modulo a prime under multiplication, the symmetries of a plane figure under composition, and sets of invertible matrices under multiplication.
A Cayley table lists every product $a*b$ with $a$ chosen from the row border and $b$ from the column border.
In a group table every element appears exactly once in each row and once in each column, so the table is a Latin square; a repeat inside a row is an immediate proof that the structure is not a group.
A group is cyclic when a single element $g$ generates it, that is $G=\{g^0,g^1,\dots,g^{n-1}\}$ where $n$ is the order of $G$ ; such a $g$ is called a generator.
Every cyclic group is abelian, but the converse is false, and the quickest way to show a group of order $n$ is not cyclic is to show that no element has order $n$ .

Worked example

Show that $\{1,2,3,4\}$ under multiplication modulo $5$ is cyclic and write down all its generators.

1.Take $g=2$ : $2^1=2$ , $2^2=4$ , $2^3=8\equiv3$ , $2^4=16\equiv1$ .
2.These four powers are the whole set, so $2$ is a generator and the group is cyclic of order $4$ .
3.Take $g=3$ : $3^1=3$ , $3^2=9\equiv4$ , $3^3=27\equiv2$ , $3^4=81\equiv1$ , so $3$ is also a generator.
4.The element $4$ has $4^2=16\equiv1$ , so its order is $2$ and it generates only $\{1,4\}$ ; the identity $1$ generates only itself.

Answer: The group is cyclic and its generators are $2$ and $3$ .

Common mistakes

Don't fall into the trap of reading a Cayley table with the row and column borders the wrong way round in a non-abelian group.
Don't fall into the trap of calling a group cyclic because some element has order greater than $1$ ; the order must equal the order of the group.
Don't fall into the trap of listing the identity as a generator of a group of order greater than $1$ .

Exam tip

To find generators, take powers of each element in turn and stop as soon as you reach the identity; the number of powers used is that element's order.

Tier 1 · Easy

ORIGINAL

1.

For the group

\{0,1,2,3\}

under addition modulo

4

, state the identity element and write down the inverse of each element.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Construct the Cayley table for

\{1,2,4\}

under multiplication modulo

7

and use it to state the order of each element.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that every cyclic group is abelian.

(4)

(Total for Question 1 is 4 marks)

1.3

Year 1

The order of a group and the order of an element. Subgroups.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

The order of a finite group $G$ , written $|G|$ , is the number of elements it contains. The order of an element $a\in G$ is the least positive integer $m$ with $a^m=e$ ; in a finite group such an $m$ always exists.
A subgroup $H$ of $G$ is a subset of $G$ that is itself a group under the same operation.
To test a subset for being a subgroup you need three things: $H$ is non-empty (it is enough to show $e\in H$ ), $H$ is closed under the operation, and $H$ is closed under taking inverses.
For a finite subset the inverse condition is automatic once closure holds, because the powers of any $a\in H$ eventually return to $e$ , and then $a^{m-1}$ is the inverse of $a$ .
Every group has the trivial subgroup $\{e\}$ and itself as subgroups; any other subgroup is called proper and non-trivial.

Worked example

Show that $H=\{1,4,11,14\}$ is a subgroup of the group of invertible residues modulo $15$ under multiplication.

1. $H$ contains the identity $1$ , so it is non-empty.
2.Closure: $4\times4=16\equiv1$ , $4\times11=44\equiv14$ , $4\times14=56\equiv11$ , $11\times11=121\equiv1$ , $11\times14=154\equiv4$ and $14\times14=196\equiv1$ .
3.Every product lies in $H$ , so $H$ is closed.
4.Inverses: $4\times4\equiv1$ , $11\times11\equiv1$ and $14\times14\equiv1$ , so each element is its own inverse and lies in $H$ .

Answer: $H$ is a subgroup of order $4$ ; every non-identity element has order $2$ .

Common mistakes

Don't fall into the trap of giving the order of an element as any exponent that returns the identity rather than the least one.
Don't fall into the trap of testing closure on only a few products instead of all of them.
Don't fall into the trap of forgetting to check that the identity of the whole group actually lies in the proposed subgroup.

Exam tip

When a question asks for the order of an element, show the successive powers; the examiner needs to see that no earlier power gave the identity.

Tier 1 · Easy

ORIGINAL

1.

Define the order of a finite group and the order of an element of a group.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the order of every element of the group

\{0,1,2,3,4,5\}

under addition modulo

6

.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Let

H

be a non-empty finite subset of a group

G

that is closed under the operation of

G

. Prove that

H

is a subgroup of

G

.

(6)

(Total for Question 1 is 6 marks)

1.4

Year 1

Lagrange's theorem.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Lagrange's theorem states that if $H$ is a subgroup of a finite group $G$ then $|H|$ divides $|G|$ . Two consequences do most of the work in examinations.
First, the order of any element divides the order of the group, because the powers of that element form a subgroup whose order is the element's order.
Second, if $|G|=n$ then $a^{n}=e$ for every $a\in G$ , since the order $m$ of $a$ divides $n$ and $a^{n}=(a^{m})^{n/m}=e$ .
A third consequence is that a group of prime order $p$ has no subgroups other than $\{e\}$ and itself, so any non-identity element generates the whole group and the group must be cyclic.
Lagrange's theorem is a one-way statement: a divisor of $|G|$ need not be the order of any subgroup in general, so never claim a subgroup exists simply because its order divides $|G|$ .

Worked example

A group $G$ has order $10$ . Write down the possible orders of the elements of $G$ and explain why $G$ must contain an element of order $2$ or an element of order $5$ .

1.By Lagrange's theorem the order of each element divides $10$ , so the possible orders are $1$ , $2$ , $5$ and $10$ .
2.Only the identity has order $1$ , so the other nine elements have order $2$ , $5$ or $10$ .
3.If some element has order $10$ then its fifth power has order $2$ .
4.Otherwise every non-identity element has order $2$ or $5$ , and there is at least one such element.

Answer: The possible orders are $1$ , $2$ , $5$ and $10$ , and in every case $G$ contains an element of order $2$ or of order $5$ .

Common mistakes

Don't fall into the trap of stating the converse and asserting that a subgroup exists for every divisor of the group order.
Don't fall into the trap of applying the theorem to an infinite group, where the counting argument does not apply.
Don't fall into the trap of confusing the order of a subgroup with the number of subgroups.

Exam tip

Write down the divisors of $|G|$ first; almost every Lagrange question is then a short elimination argument over that list.

Tier 1 · Easy

ORIGINAL

1.

State Lagrange's theorem for a finite group.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Use Lagrange's theorem to show that a group of order

7

has no subgroups other than the trivial subgroup and the group itself.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that every group of prime order

p

is cyclic and is generated by any of its non-identity elements.

(5)

(Total for Question 1 is 5 marks)

1.5

Year 2

Isomorphism.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Two groups $(G,*)$ and $(H,\circ)$ are isomorphic when there is a bijection $\varphi:G\to H$ with $\varphi(a*b)=\varphi(a)\circ\varphi(b)$ for all $a,b\in G$ ; such a $\varphi$ is an isomorphism.
Isomorphic groups have identical structure, so everything defined by the operation transfers: the identity maps to the identity, inverses map to inverses, each element has the same order as its image, and the Cayley tables agree after relabelling.
That gives the standard method for showing two groups are not isomorphic: find a structural feature they do not share, most often the list of element orders, or the fact that one is abelian and the other is not.
To show two finite groups are isomorphic, it is usually quickest to find a generator of one and map it to a generator of the other, extending by $\varphi(g^{k})=h^{k}$ .

Worked example

Show that $\{1,2,3,4\}$ under multiplication modulo $5$ is isomorphic to $\{0,1,2,3\}$ under addition modulo $4$ .

1.The first group is cyclic with generator $2$ , since $2,4,3,1$ are the successive powers of $2$ modulo $5$ .
2.The second group is cyclic with generator $1$ .
3.Define $\varphi(2^{k})=k$ for $k=0,1,2,3$ , giving $\varphi(1)=0$ , $\varphi(2)=1$ , $\varphi(4)=2$ and $\varphi(3)=3$ .
4. $\varphi$ is a bijection, and $\varphi(2^{j}\times2^{k})=\varphi(2^{j+k})=j+k$ modulo $4$ , which is $\varphi(2^{j})+\varphi(2^{k})$ .

Answer: $\varphi$ is an isomorphism, so the two groups are isomorphic.

Common mistakes

Don't fall into the trap of giving a bijection without checking that it preserves the operation.
Don't fall into the trap of claiming two groups are isomorphic merely because they have the same order.
Don't fall into the trap of comparing element orders without stating which group each order belongs to.

Exam tip

A table of element orders for both groups is the fastest disproof of isomorphism, and it earns full marks when the two lists are stated and compared.

Tier 1 · Easy

ORIGINAL

1.

Define what it means for two groups

(G,*)

and

(H,\circ)

to be isomorphic.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Show that the group

\{1,3,5,7\}

under multiplication modulo

8

is not isomorphic to the group

\{1,2,3,4\}

under multiplication modulo

5

.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Let

\varphi:G\to H

be an isomorphism of groups. Prove that

\varphi(a^{-1})=\left(\varphi(a)\right)^{-1}

for every

a\in G

.

(5)

(Total for Question 1 is 5 marks)

Groups (Further Pure 2)

The Axioms of a group.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Examples of groups. Cayley tables. Cyclic groups.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The order of a group and the order of an element. Subgroups.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Lagrange's theorem.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Isomorphism.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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