## 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

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]

[N/A]

[N/A]