# IB DP Maths Topic 8.12 Definition of a group homomorphism. HL Paper 3

## Question

Let $$f:G \to H$$ be a homomorphism of finite groups.

Prove that $$f({e_G}) = {e_H}$$, where $${e_G}$$ is the identity element in $$G$$ and $${e_H}$$ is the identity

element in $$H$$.


a.

(i)     Prove that the kernel of $$f,{\text{ }}K = {\text{Ker}}(f)$$, is closed under the group operation.

(ii)     Deduce that $$K$$ is a subgroup of $$G$$.


b.

(i)     Prove that $$gk{g^{ – 1}} \in K$$ for all $$g \in G,{\text{ }}k \in K$$.

(ii)     Deduce that each left coset of K in G is also a right coset.


c.

## Markscheme

$$f(g) = f({e_G}g) = f({e_G})f(g)$$ for $$g \in G$$     M1A1

$$\Rightarrow f({e_G}) = {e_H}$$     AG

[2 marks]

a.

(i)     closure: let $${k_1}$$ and $${k_2} \in K$$, then $$f({k_1}{k_2}) = f({k_1})f({k_2})$$     M1A1

$$= {e_H}{e_H} = {e_H}$$     A1

hence $${k_1}{k_2} \in K$$     R1

(ii)     K is non-empty because $${e_G}$$ belongs to K     R1

a closed non-empty subset of a finite group is a subgroup     R1AG

[6 marks]

b.

(i)     $$f(gk{g^{ – 1}}) = f(g)f(k)f({g^{ – 1}})$$     M1

$$= f(g){e_H}f({g^{ – 1}}) = f(g{g^{ – 1}})$$     A1

$$= f({e_G}) = {e_H}$$     A1

$$\Rightarrow gk{g^{ – 1}} \in K$$     AG

(ii)     clear definition of both left and right cosets, seen somewhere.     A1

use of part (i) to show $$gK \subseteq Kg$$     M1

similarly $$Kg \subseteq gK$$     A1

hence $$gK = Kg$$     AG

[6 marks]

c.

[N/A]

a.

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b.

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c.

## Question

The group $$\{ G,{\rm{ }} * {\rm{\} }}$$ has identity $${e_G}$$ and the group $$\{ H,{\text{ }} \circ \}$$ has identity $${e_H}$$. A homomorphism $$f$$ is such that $$f:G \to H$$. It is given that $$f({e_G}) = {e_H}$$.

Prove that for all $$a \in G,{\text{ }}f({a^{ – 1}}) = {\left( {f(a)} \right)^{ – 1}}$$.


a.

Let $$\{ H,{\text{ }} \circ \}$$ be the cyclic group of order seven, and let $$p$$ be a generator.

Let $$x \in G$$ such that $$f(x) = {p^{\text{2}}}$$.

Find $$f({x^{ – 1}})$$.


b.

Given that $$f(x * y) = p$$, find $$f(y)$$.


c.

## Markscheme

$$f({e_G}) = {e_H} \Rightarrow f(a * {a^{ – 1}}) = {e_H}$$     M1

$$f$$ is a homomorphism so $$f(a * {a^{ – 1}}) = f(a) \circ f({a^{ – 1}}) = {e_H}$$     M1A1

by definition $$f(a) \circ {\left( {f(a)} \right)^{ – 1}} = {e_H}$$ so $$f({a^{ – 1}}) = {\left( {f(a)} \right)^{ – 1}}$$ (by the left-cancellation law)     R1

[4 marks]

a.

from (a) $$f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}$$

hence $$f({x^{ – 1}}) = {({p^2})^{ – 1}} = {p^5}$$     M1A1

[2 marks]

b.

$$f(x * y) = f(x) \circ f(y)\;\;\;$$(homomorphism)     (M1)

$${p^2} \circ f(y) = p$$     A1

$$f(y) = {p^5} \circ p$$     (M1)

$$= {p^6}$$     A1

[4 marks]

Total [10 marks]

c.

## Examiners report

Part (a) was well answered by those who understood what a homomorphism is. However many candidates simply did not have this knowledge and consequently could not get into the question.

a.

Part (b) was well answered, even by those who could not do (a). However, there were many who having not understood what a homomorphism is, made no attempt on this easy question part. Understandably many lost a mark through not simplifying $${p^{ – 2}}$$ to $${p^5}$$.

b.

Those who knew what a homomorphism is generally obtained good marks in part (c).

c.

## Question

A group $$\{ D,{\text{ }}{ \times _3}\}$$ is defined so that $$D = \{ 1,{\text{ }}2\}$$ and $${ \times _3}$$ is multiplication modulo $$3$$.

A function $$f:\mathbb{Z} \to D$$ is defined as $$f:x \mapsto \left\{ {\begin{array}{*{20}{c}} {1,{\text{ }}x{\text{ is even}}} \\ {2,{\text{ }}x{\text{ is odd}}} \end{array}} \right.$$.

Prove that the function $$f$$ is a homomorphism from the group $$\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\}$$.


a.

Find the kernel of $$f$$.


b.

Prove that $$\{ {\text{Ker}}(f),{\text{ }} + \}$$ is a subgroup of $$\{ \mathbb{Z},{\text{ }} + \}$$.


c.

## Markscheme

consider the cases, $$a$$ and $$b$$ both even, one is even and one is odd and $$a$$ and $$b$$ are both odd     (M1)

calculating $$f(a + b)$$ and $$f(a){ \times _3}f(b)$$ in at least one case     M1

if $$a$$ is even and $$b$$  is even, then $$a + b$$ is even

so$$\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}1 = 1$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

if one is even and the other is odd, then $$a + b$$ is odd

so$$\;\;\;f(a + b) = 2.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}2 = 2$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

if $$a$$ is odd and $$b$$ is odd, then $$a + b$$ is even

so$$\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 2{ \times _3}2 = 1$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

as$$\;\;\;f(a + b) = f(a){ \times _3}f(b)\;\;\;$$in all cases, so$$\;\;\;f:\mathbb{Z} \to D$$ is a homomorphism     R1AG

[6 marks]

a.

$$1$$ is the identity of $$\{ D,{\text{ }}{ \times _3}\}$$     (M1)(A1)

so$$\;\;\;{\text{Ker}}(f)$$ is all even numbers     A1

[3 marks]

b.

METHOD 1

sum of any two even numbers is even so closure applies     A1

associative as it is a subset of $$\{ \mathbb{Z},{\text{ }} + \}$$     A1

identity is $$0$$, which is in the kernel     A1

the inverse of any even number is also even     A1

METHOD 2

$${\text{ker}}(f) \ne \emptyset$$

$${b^{ – 1}} \in {\text{ker}}(f)$$ for any $$b$$

$$a{b^{ – 1}} \in {\text{ker}}(f)$$ for any $$a$$ and $$b$$

Note:     Allow a general proof that the Kernel is always a subgroup.

[4 marks]

Total [13 marks]

c.

[N/A]

a.

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b.

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c.

## Question

The group $$\{ G,{\text{ }} * \}$$ is Abelian and the bijection $$f:{\text{ }}G \to G$$ is defined by $$f(x) = {x^{ – 1}},{\text{ }}x \in G$$.

Show that $$f$$ is an isomorphism.

## Markscheme

we need to show that $$f(a * b) = f(a) * f(b)$$     R1

Note:     This R1 may be awarded at any stage.

let $$a,{\text{ }}b \in G$$     (M1)

consider $$f(a) * f(b)$$     M1

$$= {a^{ – 1}} * {b^{ – 1}}$$    A1

consider $$f(a * b) = {(a * b)^{ – 1}}$$     M1

$$= {b^{ – 1}} * {a^{ – 1}}$$    A1

$$= {a^{ – 1}} * {b^{ – 1}}$$ since $$G$$ is Abelian     R1

hence $$f$$ is an isomorphism     AG

[7 marks]

## Examiners report

A surprising number of candidates wasted time and unrewarded effort showing that the mapping $$f$$, stated to be a bijection in the question, actually was a bijection. Many candidates failed to get full marks by not properly using the fact that the group was stated to be Abelian. There were also candidates who drew the graph of $$y = \frac{1}{x}$$ or otherwise assumed that the inverse of $$x$$ was its reciprocal – this is unacceptable in the context of an abstract group question.

## Question

The set of all permutations of the list of the integers 1, 2, 3  4 is a group, S4, under the operation of function composition.

In the group S4 let $${p_1} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&3&1&4 \end{array} \hfill \\ \end{gathered} \right)$$ and $${p_2} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&1&3&4 \end{array} \hfill \\ \end{gathered} \right)$$.

Determine the order of S4.


a.

Find the proper subgroup H of order 6 containing $${p_1}$$, $${p_2}$$ and their compositions. Express each element of H in cycle form.


b.

Let $$f{\text{:}}\,{S_4} \to {S_4}$$ be defined by $$f\left( p \right) = p \circ p$$ for $$p \in {S_4}$$.

Using $${p_1}$$ and $${p_2}$$, explain why $$f$$ is not a homomorphism.


c.

## Markscheme

number of possible permutations is 4 × 3 × 2 × 1       (M1)

= 24(= 4!)      A1

[2 marks]

a.

attempting to find one of $${p_1} \circ {p_1}$$, $${p_1} \circ {p_2}$$ or $${p_2} \circ {p_1}$$     M1

$${p_1} \circ {p_1} = \left( {132} \right)$$ or equivalent (eg, $${p_1}^{ – 1} = \left( {132} \right)$$)    A1

$${p_1} \circ {p_2} = \left( {13} \right)$$ or equivalent (eg, $${p_2} \circ {p_1} \circ {p_1} = \left( {13} \right)$$)    A1

$${p_2} \circ {p_1} = \left( {23} \right)$$ or equivalent (eg, $${p_1} \circ {p_1} \circ {p_2} = \left( {23} \right)$$)    A1

Note: Award A1A0A0 for one correct permutation in any form; A1A1A0 for two correct permutations in any form.

$$e = \left( 1 \right)$$, $${p_1} = \left( {123} \right)$$ and $${p_2} = \left( {12} \right)$$     A1

Note: Condone omission of identity in cycle form as long as it is clear it is considered one of the elements of H.

[5 marks]

b.

METHOD 1

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to express one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$ in terms of $${p_1}$$ and $${p_2}$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = {p_1} \circ {p_2} \circ {p_1} \circ {p_2}$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = {p_1} \circ {p_1} \circ {p_2} \circ {p_2}$$     A1

$$\Rightarrow {p_2} \circ {p_1} = {p_1} \circ {p_2}$$     A1

but $${p_1} \circ {p_2} \ne {p_2} \circ {p_1}$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

METHOD 2

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to find one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = e$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = \left( {132} \right)$$     (M1)A1

so $$f\left( {{p_1} \circ {p_2}} \right) \ne f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

[5 marks]

c.

[N/A]

a.

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b.

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c.

## Question

The groups $$\{ K,{\text{ }} * \}$$ and $$\{ H,{\text{ }} \odot \}$$ are defined by the following Cayley tables.

G H By considering a suitable function from G to H , show that a surjective homomorphism exists between these two groups. State the kernel of this homomorphism.

## Markscheme

consider the function f given by

$$f(E) = e$$

$$f(A) = e$$

$$f(B) = a$$     M1A1

$$f(C) = a$$

then, it has to be shown that

$$f(X * Y) = f(X) \odot f(Y){\text{ for all }}X{\text{ , }}Y \in G$$     (M1)

consider

$$f\left( {(E{\text{ or }}A) * (E{\text{ or }}A)} \right) = f(E{\text{ or }}A) = e;{\text{ }}f(E{\text{ or }}A) \odot f(E{\text{ or }}A) = e \odot e = e$$     M1A1

$$f\left( {(E{\text{ or }}A) * (B{\text{ or }}C)} \right) = f(B{\text{ or }}C) = a;{\text{ }}f(E{\text{ or }}A) \odot f(B{\text{ or }}C) = e \odot a = a$$     A1

$$f\left( {(B{\text{ or }}C) * (B{\text{ or }}C)} \right) = f(E{\text{ or }}A) = e;{\text{ }}f(B{\text{ or }}C) \odot f(B{\text{ or }}C) = a \odot a = e$$     A1

since the groups are Abelian, there is no need to consider $$f\left( {(B{\text{ or }}C) * (E{\text{ or }}A)} \right)$$     R1

the required property is satisfied in all cases so the homomorphism exists

Note: A comprehensive proof using tables is acceptable.

the kernel is $$\{ E,{\text{ }}A\}$$     A1

[9 marks]

[N/A]