IB DP Maths Topic 8.12 Definition of the kernel of a 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 $$H$$.

[2]
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 $$G$$.

[6]
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.

[6]
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.

[N/A]

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 $$\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\}$$.

[6]
a.

Find the kernel of $$f$$.

[3]
b.

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

[4]
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.

[N/A]

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.

[2]
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.

[5]
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.

[5]
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.

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a.

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b.

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c.