## 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 \(H\).

(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 \(G\).

(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.

**Answer/Explanation**

## 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]*

(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]*

(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]*

## Examiners report

[N/A]

[N/A]

[N/A]

## 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 \(a \in G,{\text{ }}f({a^{ – 1}}) = {\left( {f(a)} \right)^{ – 1}}\).

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 \(f({x^{ – 1}})\).

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

**Answer/Explanation**

## 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]*

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

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

*[2 marks]*

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

\({p^2} \circ f(y) = p\) *A1*

\(f(y) = {p^5} \circ p\) *(M1)*

\( = {p^6}\) *A1*

*[4 marks]*

*Total [10 marks]*

## 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.

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 \({p^5}\).

Those who knew what a homomorphism is generally obtained good marks in part (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 \(\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\} \).

Find the kernel of \(f\).

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

**Answer/Explanation**

## 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]*

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

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

*[3 marks]*

**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]**

## Examiners report

[N/A]

[N/A]

[N/A]

## 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.

**Answer/Explanation**

## 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, *S*_{4}, under the operation of function composition.

In the group *S*_{4} 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 *S*_{4}.

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

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.

**Answer/Explanation**

## Markscheme

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

= 24(= 4!) ** A1**

**[2 marks]**

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;

*for two correct permutations in any form.*

**A1A1A0**\(e = \left( 1 \right)\), \({p_1} = \left( {123} \right)\) 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]**

**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

*is awarded.*

**M1****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

*is awarded.*

**M1****Note:** Award marks only if \({p_1}\) and \({p_2}\) are used; cycle form is not required.

**[5 marks]**

## Examiners report

[N/A]

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## 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.

**Answer/Explanation**

## 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]*